Determine Where the Function is Continuous Rational Irrational

Integration of Rational and Irrational Functions: Methods, Formula

Integration of rational and irrational functions: We are familiar with functions and their derivatives. We often are not given an expression for a function, \(f\left( x \right)\). We only have an expression for the rate of change of that function, \(f'\left( x \right)\). When this happens, to find the original function, we would need to find antiderivatives and to do that we will need to use integration.

We can find an expression for velocity by differentiating the expression for displacement \(v = \frac{{ds}}{{dt}}\). Since integration is the inverse process of differentiation, to obtain the displacement,\(s\) of an object at time \(t\) for the given expression for velocity \(v\) we would use \(s = \int v dt\). Similarly, with the help of integration, the velocity of an object at time \(t\) with acceleration \(a\), can be calculated by \(v = \int a dt\).

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What are Partial Fractions?

A rational function has two or more rational expressions into one. Here, when one rational expression is given, we express it as the sum of two or more rational expressions. A special type of sum of simpler fractions is called partial fraction decomposition. Each term in the sum is a partial fraction.

Types of Partial Fractions

There are four types of partial fraction and their integration:

Type	Expression	Integration
(i)	\(\frac{A}{{x – a}}\)	\(\int {\frac{A}{{x – a}}} dx = A\ln |x – a| + C\)
(ii)	\(\frac{A}{{{{(x – a)}^k}}}\) Where, \(k\) is a positive integer \( \ge 2\)	\(\int {\frac{A}{{{{(x – a)}^k}}}} dx = A\int {{{(x – a)}^{ – k}}} dx = \frac{{A{{(x – a)}^{ – k + 1}}}}{{ – k + 1}} + C\)
(iii)	\(\frac{{Ax + B}}{{{x^2} + px + q}}\) Here, the roots of the denominator are imaginary, i.e. \({p^2} – 4q < 0\)	\(\int {\frac{{Ax + B}}{{{x^2} + px + q}}} dx\) \( = \frac{A}{2}\ln \left| {{x^2} + px + q} \right| + \frac{{2B – Ap}}{{\sqrt {4q – {p^2}} }}{\tan ^{ – 1}}\frac{{2x – p}}{{\sqrt {4q – {p^2}} }} + C\)
(iv)	\(\frac{{Ax + B}}{{{{\left( {{x^2} + px + q} \right)}^k}}}\) Where, \(k\) is a positive integer \( \ge 2\), the roots of the denominator are imaginary.	\(\int {\frac{{Ax + B}}{{{{\left( {{x^2} + px + q} \right)}^k}}}} dx = \int {\frac{{\frac{A}{2}(2x + p) + \left( {B – \frac{{Ap}}{2}} \right)}}{{{{\left( {{x^2} + px + q} \right)}^k}}}} dx\) \( = \frac{A}{2}\int {\frac{{2x + p}}{{{{\left( {{x^2} + px + q} \right)}^k}}}} dx + \left( {B – \frac{{Ap}}{2}} \right)\int {\frac{{dx}}{{{{\left( {{x^2} + px + q} \right)}^k}}}} \) \( = \frac{1}{{(k – 1){{\left( {{x^2} + px + q} \right)}^{k – 1}}}} + \left( {B – \frac{{Ap}}{2}} \right)\int {\frac{{dx}}{{{{\left( {{x^2} + px + q} \right)}^k}}}} \)

Note that every proper rational fraction may be represented as a sum of partial fractions.

Consider a rational function, \(f(x) = \frac{{P(x)}}{{Q(x)}}\), where \(P(x)\) and \(Q(x)\) are polynomials.

It is possible to express \(f(x)\) as a sum of simpler fractions if the degree of \(P\) is less than the degree of \(Q\). Simply put, \(f\) is a proper rational function.

Steps to Form Partial Fractions

Use the following steps to form a partial fraction decomposition of a rational expression.

Step 1: If \(f\) is improper, i.e. \(\deg(P) \ge \deg (Q)\), then divide \(P\) by \(Q\) until a remainder \(R(x)\) is obtained such that \(\deg (R) < \deg (Q)\).

\(\therefore \,\frac{{P(x)}}{{Q(x)}} = S(x) + \frac{{R(x)}}{{Q(x)}}\)

Here, \(S\) and \(R\) are also polynomials.

Step 2: Factor \(Q(x)\) into factors of the form \({(ax + b)^m}\) or \({\left( {c{x^2} + dx + e} \right)^n}\), where \(c{x^2} + dx + e\) is irreducible, and \(m\) and \(n\) are integers.

Note:

• Any polynomial \(Q\) can be factorised as a product of linear factors of the form \(ax + b\) and irreducible quadratic factors of the form \(a{x^2} + bx + c\), where \({b^2} – 4ac < 0\).
• If the equation \(Q(x) = 0\) cannot be solved algebraically, then the method of partial fractions naturally fails, and we must use the other methods.

Step 3:

For each type of linear factor listed below, the decomposition must include the corresponding term given in the table.

Form of the Linear Factor	Decomposition
\((ax + b)\)	\(\frac{A}{{ax + b}}\)
\({(ax + b)^m}\)	\(\frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{(ax + b)}^2}}} + \ldots + \frac{{{A_m}}}{{{{(ax + b)}^m}}}\)
\(\left( {c{x^2} + dx + e} \right)\)	\(\frac{{Bx + C}}{{cx + dx + e}}\)
\({\left( {c{x^2} + dx + e} \right)^n}\)	\(\frac{{{B_1}x + {C_1}}}{{c{x^2} + dx + e}} + \frac{{{B_2}x + {C_2}}}{{{{\left( {c{x^2} + dx + e} \right)}^2}}} + \ldots + \frac{{{B_n} + {C_n}}}{{{{\left( {c{x^2} + dx + e} \right)}^n}}}\)

Step 4: Use techniques to solve for the constants in the numerators. There are four possible cases here.

Case 1: \(Q\left( x \right)\) is a product of distinct linear factors

Write \(Q(x) = \left( {{a_1}x + {b_1}} \right)\left( {{a_2}x + {b_2}} \right) \ldots \left( {{a_k}x + {b_k}} \right)\), where no factor is repeated.

In this case, the partial fraction theorem states that there exist constants \({A_1},\,{A_2},\, \ldots {A_k}\) such that

\(\frac{{R(x)}}{{Q(x)}} = \frac{{{A_1}}}{{{a_1}x + {b_1}}} + \frac{{{A_2}}}{{{a_2}x + {b_2}}} + \ldots + \frac{{{A_k}}}{{{a_k}x + {b_k}}}\)

Find the constants \({A_1},\,{A_2},\, \ldots {A_k}\), and then integrate both sides.

Case 2: \(Q\left( x \right)\) is a product of linear factors, some of which are repeated

Suppose the first linear factor \(\left( {{a_1}x + {b_1}} \right)\) is repeated \(r\) times, i.e. \({\left( {{a_1}x + {b_1}} \right)^r}\) occurs in the factorisation of \(Q\left( x \right)\), then instead of a single term \(\frac{{{A_1}}}{{{a_1}x + {b_1}}}\), we will use \(r\) terms.

\(\frac{{P(x)}}{{{{(x – a)}^r}Q(x)}} = \frac{{{A_1}}}{{{a_1}x + {b_1}}} + \frac{{{A_2}}}{{{{\left( {{a_1}x + {b_1}} \right)}^2}}} + \ldots + \frac{{{A_r}}}{{{{\left( {{a_1}x + {b_1}} \right)}^r}}}\,….(i)\)

Example:

\(\frac{{{x^3} – x + 1}}{{{x^2}{{(x – 1)}^3}}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x – 1}} + \frac{D}{{{{(x – 1)}^2}}} + \frac{E}{{{{(x – 1)}^2}}}\)

• To obtain the constant of the partial fraction corresponding to \((x – a)\) multiply both sides of the identity \((i)\) by \((x – a)\) and then let \(x \to \infty \).
• To obtain the constant of the partial fraction containing \({(x – a)^r}\) in the denominator, put \(x = a\) in the given fraction except for \({(x – a)^r}\) in the denominator. This gives \({A_r} = \frac{{P(a)}}{{Q(a)}}\)
• To obtain the other constants, we multiply both sides of the identity again by \((x – a)\) and then let \(x \to \infty \). Alternatively, put some values of \(x\) (say \(0,\,1,\, – 1\)) except the roots of the denominator in the identity, thus forming a system of linear equations and solving it.

Case 3: \(Q\left( x \right)\) contains irreducible quadratic factors, none of which is repeated

If \(Q\left( x \right)\) has the factor \(\left( {a{x^2} + bx + c} \right)\)where \({b^2} – 4ac < 0\), then in addition to the partial fractions taken earlier, the expression will have a term of the form

\(\frac{{Ax + B}}{{a{x^2} + bx + c}}\)

Where, \(A\) and \(B\) are constants.

This term can be integrated by completing the square and using the formula

\(\int {\frac{{dx}}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ – 1}}\left( {\frac{x}{a}} \right) + C\)

Case 4: \(Q\left( x \right)\) contains a repeated irreducible quadratic factor

If \(Q\left( x \right)\) has the factor \({\left( {a{x^2} + bx + c} \right)^r}\), where \({b^2} – 4ac < 0\), then instead of the single partial fraction, a sum occurs in the partial fraction decomposition as shown below.

\(\frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + \ldots + \frac{{{A_r}x + {B_r}}}{{{{\left( {a{x^2} + bx + c} \right)}^r}}}\)

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Integration of Rational Functions

Sometimes partial fractions can be avoided to integrate a rational function, and we can use other methods to integrate it.

1. Integrals of the form \(\int {\frac{{p{x^2} + q}}{{\left( {{x^2} + a} \right)\left( {{x^2} + b} \right)}}} dx\)

Assume \({x^2} = t\), to find partial fractions.

Suppose that \(\frac{{pt + q}}{{(t + a)(t + b)}} = \frac{A}{{t + a}} + \frac{B}{{t + b}}\) then the integral becomes \(\int {\frac{A}{{\left( {{x^2} + a} \right)}}} dx + \int {\frac{B}{{{x^2} + b}}} dx\)

2. Integrals of the form \(\int {\frac{{P(x)}}{{Q(x)}}} dx\) where \(Q(x)\) has a linear factor with high index

Substitute the linear factors \( = \frac{1}{t}\).

3. Substitution

(i) Integrals of the form \(\int {\frac{{{x^m}}}{{{{(ax + b)}^n}}}} dx,\,m,\,n \in {\Bbb N}\)

Put \(ax + b = t\)

The integral becomes \(\frac{1}{{{a^{m + 1}}}}\int {\frac{{{{(t – b)}^m}dt}}{{{t^n}}}} \).

The required integral can be immediately obtained by expanding using the binomial theorem and integrating each term separately.

Note: By the same process, we can integrate \(\int {\frac{{{{(a + bx)}^m}}}{{{{\left( {{a^\prime } + {b^\prime }x} \right)}^n}}}} dx\), where \(m\) is a positive integer, and \(n\) is a rational number.

Form of Integrals	Substitution
\(\int {\frac{{{x^m}}}{{{{(ax + b)}^n}}}} dx,\,m,\,n \in {\Bbb N}\)	\(ax + b = t\)
\(\int {\frac{{{{(a + bx)}^m}}}{{{{\left( {{a^\prime } + {b^\prime }x} \right)}^n}}}} dx\) Where, \(m\) is a positive integer, and \(n\) is a rational number	\(ax + b = t\)
\(\int {\frac{{dx}}{{{x^m}{{(ax + b)}^n}}}} ,\,m,\,n \in {\Bbb N}\)	\(\frac{{ax + b}}{x} = t\)
\(\int {\frac{{dx}}{{{{(x – a)}^m}{{(x – b)}^n}}}} ,\,m,\,n \in {\Bbb N}\)	\(\frac{{x – a}}{{x – b}} = t\) if \(m < n\)
\(\int {\frac{{dx}}{{x\left( {a + b{x^n}} \right)}}} \)	Here, the substitution of \({x^n} = \frac{1}{t}\) , gives \( – \frac{1}{n}\int {\frac{{dt}}{{at + b}}} = – \frac{1}{n}\ln |at + b| + C\) \( = \frac{1}{{na}}\ln \left( {\frac{{{x^n}}}{{a + b{x^n}}}} \right) + C\)
\(\int {\frac{{{x^{2m + 1}}}}{{{{\left( {a{x^2} + b} \right)}^n}}}} dx,\,m,\,n \in {\Bbb Z}\)	Put \(a{x^2} + b = t\) The integral becomes \(\int {\frac{{{{(t – b)}^m}}}{{2{a^{m + 1}}{t^n}}}} dt\), which is immediately integrable by the aid of the Binomial Theorem.

4. Integration By Parts: Suppose \(\int f gdx\) is given where \(f,\,g\) are differentiable functions. Here, we call \(f\) as the first function and \(g\) as the second function to easily obtain the following formula.

\(\int f gdx = f\left[ {\int g dx} \right] – \int {{f^\prime }} \left[ {\int g dx} \right]dx\)

Now, how can we determine which is the first function and which is the second?

To know this, use the following strategies.

1. When the integrand contains only one function which is not directly integrable such as \(\ln x,\,{\sin ^{ – 1}}x,\,{\cos ^{ – 1}}x\) and \({\tan ^{ – 1}}\,x\), we consider unity as the second function and the given integrand as the first function.
2. Suppose the integrand is a product of two functions, such that one is directly integrable and the other is not. Then, we consider the direct integrable function as the second function, while the other is the first function.
3. If integrand is product of two functions, where both are directly integrable, then the first function is chosen in the order "ILATE" as listed below.
   • Inverse trigonometric
   • Logarithmic
   • Algebraic
   • Trigonometric
   • Exponential

Note:

1. There are several integrals which are not integrable. Like \(\int {{e^{{x^2}}}} dx,\,\int {\sqrt {\sin x} }\, dx,\,\int {\ln } \sin x\,dx,\,\int {\frac{1}{{\sqrt[3]{{1 + {x^2}}}}}} dx\) etc.
2. \(\int {{e^x}} \left( {f(x) + {f^\prime }(x)} \right)dx = {e^x}f(x) + C\)

Integration of Irrational Functions

The concept is explained in detail below for a better understanding.

Integration by Rationalisation

Certain types of integrals of algebraic irrational expressions can be reduced to integrals of rational functions by an appropriate change of the variable. Such transformation of an integral is called its rationalisation.

1. Let \(n\) be a positive integer. Any rational function of \(x\) and \(\sqrt[n]{{ax + b}}\) can be transformed into a rational function by the substitution \({t^n} = ax + b\) and thus can be integrated by partial fractions.
2. Integrals of the form \(\int {\frac{{dx}}{{x\sqrt {a{x^n} + b} }}} \)
   Here substitute \(a{x^n} + b = {t^2}\)
3. Integrals of the form \(\int {\frac{{dx}}{{{{\left( {a + c{x^2}} \right)}^{\frac{3}{2}}}}}} \)
   Substitute \(x = \frac{1}{t}\), and the integral becomes
   \(\int – \frac{{tdt}}{{{{\left( {a{t^2} + c} \right)}^{\frac{3}{2}}}}} = \frac{1}{{a{{\left( {a{t^2} + c} \right)}^{\frac{1}{2}}}}} + C\)
   \(\therefore \,\int {\frac{{dx}}{{{{\left( {a + c{x^2}} \right)}^{\frac{3}{2}}}}}} = \frac{x}{{a{{\left( {a + c{x^2}} \right)}^{\frac{1}{2}}}}} + C\)
4. Integrals of the form \(\int {\frac{{xdx}}{{{{\left( {a + 2bx + c{x^2}} \right)}^{\frac{3}{2}}}}}} \)
   Substitute \(x = \frac{1}{t}\). Then, the integral becomes
   \(\int – \frac{{dx}}{{{{\left( {a{z^2} + 2bz + c} \right)}^{\frac{3}{2}}}}}\)
   Hence, \(\int {\frac{{xdx}}{{{{\left( {a + 2bx + c{x^2}} \right)}^{\frac{3}{2}}}}}} = – \frac{{a + bx}}{{\left( {ac – {b^2}} \right){{\left( {a + 2bx + c{x^2}} \right)}^{\frac{1}{2}}}}} + C\)
5. If the integrand is a rational function of fractional powers of \(x\) i.e. the function \(R\left( {x,\,{x^{\frac{{{p_1}}}{{{q_1}}}}},\, \ldots ,\,{x^{\frac{{{p_k}}}{{{q_k}}}}}} \right)\) then the integral can be rationalised by the substitution
   \(x = {t^m}\)
   where \(m\) is the LCM of the denominators \({q_1},\,{q_2},\, \ldots ,\,{q_k}\) of several fractional powers. This means the integration of such expressions is reduced to that of rational functions.
6. For any algebraic expression containing integral powers of \(x\) along with fractional powers of an expression of the form, \(a + bx\) is immediately reduced to rational functions by the substitution \(a + bx = {t^m}\), where \(m\) is the LCM of the denominators of the several fractional powers.
7. If the integrand is a rational function of \(x\) and fractional powers of a linear fractional function of the form, \(\frac{{ax + b}}{{cx + d}}\) then rationalisation of the integral is affected by the substitution \(\frac{{cx + d}}{{ax + b}} = {t^m}\), where \(m\) is the LCM of the denominators of the several fractional powers.

Note: The integrals of the form \(\int {\frac{{dx}}{{\sqrt[n]{{{{(x – a)}^p}{{(x – b)}^q}}}}}} \) where \(p + q = 2n\) are solved more comfortably using the substitution \(\frac{{x – a}}{{x – b}} = t\).

Solved Examples on Integration of Rational and Irrational Functions

Below are a few solved examples that can help in getting a better idea.

Q.1. Evaluate \(\int {\frac{{dx}}{{\left( {{x^2} – 1} \right)(3 + 2x)}}} \)
Ans: Let \(I = \int {\frac{{dx}}{{\left( {{x^2} – 1} \right)(3 + 2x)}}} = \int {\frac{{dx}}{{(x – 1)(x + 1)(3 + 2x)}}} \)
Now let, \(\frac{1}{{(x + 1)(x – 1)(3 + 2x)}} = \frac{A}{{x + 1}} + \frac{B}{{x – 1}} + \frac{C}{{2x + 3}}\)
Or \(1 = A(x – 1)(2x + 3) + B(x + 1)(2x + 3) + C(x + 1)(x – 1)\)
Putting \(x = 1,\, – 1,\, – \frac{3}{2}\) successively, we get
\(1 = B \cdot 2 \cdot 5,\,1 = A( – 2) \cdot 1,\,1 = C \cdot \left( { – \frac{1}{2}} \right)\left( { – \frac{5}{2}} \right)\)
Therefore, \(B = \frac{1}{{10}},\,A = – \frac{1}{2},\,C = \frac{4}{5}\)
\(\therefore \,I = \int {\left( {\frac{A}{{x + 1}} + \frac{B}{{x – 1}} + \frac{C}{{2x + 3}}} \right)} dx = – \frac{1}{2}\ln |x + 1| + \frac{1}{{10}}\ln |x – 1| + \frac{2}{5}\ln |2x + 3| + k\)

Q.2. Evaluate \(\int {\frac{{1 – x + 2{x^2} – {x^3}}}{{x{{\left( {{x^2} + 1} \right)}^2}}}} dx\)
Ans: The form of the partial fraction decomposition is
\(\frac{{1 – x + 2{x^2} – {x^3}}}{{x{{\left( {{x^2} + 1} \right)}^2}}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} + \frac{{Dx + E}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)
Multiplying by \(x{\left( {{x^2} + 1} \right)^2}\), we get
\( – {x^3} + 2{x^2} – x + 1 = A{\left( {{x^2} + 1} \right)^2} + (Bx + C)x\left( {{x^2} + 1} \right) + (Dx + E)x\)
\( = A\left( {{x^4} + 2{x^2} + 1} \right) + B\left( {{x^4} + {x^2}} \right) + C\left( {{x^3} + x} \right) + D{x^2} + Ex\)
\( = (A + B){x^4} + C{x^3} + (2A + B + D){x^2} + (C + E)x + A\)
If we equate coefficients, we get the system
\(A + B = 0,\,C = – 1,\,2A + B + D = 2,\,C + E = – 1,\,A = 1\)
The solution is \(A = 1,\,B = – 1,\,C = – 1,\,D = 1\) and \(E = 0\)
Thus
\(\int {\frac{{1 – x + 2{x^2} – {x^3}}}{{x{{\left( {{x^2} + 1} \right)}^2}}}} dx = \int {\left( {\frac{1}{x} – \frac{{x + 1}}{{{x^2} + 1}} + \frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)} dx\)
\( = \int {\frac{{dx}}{x}} – \int {\frac{x}{{{x^2} + 1}}} dx – \int {\frac{{dx}}{{{x^2} + 1}}} + \int {\frac{{xdx}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \)
\( = \ln |x| – \frac{1}{2}\ln \left( {{x^2} + 1} \right) – {\tan ^{ – 1}}\,x – \frac{1}{{2\left( {{x^2} + 1} \right)}} + k\)
Hence, \(\int {\frac{{1 – x + 2{x^2} – {x^3}}}{{x{{\left( {{x^2} + 1} \right)}^2}}}} dx = \ln |x| – \frac{1}{2}\ln \left( {{x^2} + 1} \right) – {\tan ^{ – 1}}\,x – \frac{1}{{2\left( {{x^2} + 1} \right)}} + k\)

Q.3. Evaluate \(I = \int {\frac{{dx}}{{{x^2}{{(x + 2)}^3}}}} \)
Ans: \(I = \int {\frac{{dx}}{{{x^5}{{\left( {\frac{{x + 2}}{x}} \right)}^3}}}} \)
Put \(\frac{{x + 2}}{x} = t \Rightarrow 1 + \frac{2}{x} = t \Rightarrow – \frac{2}{{{x^2}}}dx = dt\)
\(I = – \frac{1}{2}\int {\frac{{{{(t – 1)}^3}}}{{8{t^3}}}} dt = – \frac{1}{{16}}\int {\frac{{{t^3} – 3{t^2} + 3t – 1}}{{{t^3}}}} dt\)
\( = – \frac{1}{{16}}\left[ {\int 1 dt – \int {\frac{3}{t}} dt + \int {\frac{3}{{{t^2}}}} dt – \int {\frac{1}{{{t^3}}}} dt} \right]\)
\( = – \frac{1}{{16}}\left[ {t – 3\ln |t| – \frac{3}{t} + \frac{1}{{2{t^2}}}} \right] + k\)
Put the value of \(t = \frac{{x + 2}}{x}\), we get
\(\int {\frac{{dx}}{{{x^2}{{(x + 2)}^3}}}} = – \frac{{x + 2}}{{16x}} + \frac{3}{{16}}\ln \left( {\frac{{x + 2}}{x}} \right) + \frac{3}{{16}}\left( {\frac{x}{{x + 2}}} \right) – \frac{1}{{32}}\left( {\frac{{{x^2}}}{{{{(x + 2)}^2}}}} \right) + k\)

Q.4. Evaluate \(\int {\frac{{dx}}{{\sqrt[3]{{x + 1}} + \sqrt {x + 1} }}} \)
Ans: Let \(I = \int {\frac{{dx}}{{\sqrt[3]{{x + 1}} + \sqrt {x + 1} }}} \)
\( \Rightarrow I = \int {\frac{{dx}}{{{{(x + 1)}^{\frac{1}{3}}} + {{(x + 1)}^{\frac{1}{2}}}}}} \)
The least common multiple of \(2\) and \(3\) is \(6\). So, substitute \(x + 1 = {t^6}\)
\( \Rightarrow dx = 6{x^5}dt\)
\( \Rightarrow I = \int {\frac{{6{t^5}dt}}{{{t^2} + {t^3}}}} = 6\int {\frac{{{t^3}dt}}{{1 + t}}} \)
\( \Rightarrow I = 6\int {\left( {{t^2} – t + 1 – \frac{1}{{1 + t}}} \right)} dt\)
\( \Rightarrow I = 6\left( {\frac{{{t^3}}}{3} – \frac{{{t^2}}}{2} + t – \ln |t + 1|} \right) + C\)
On substituting \(t = {(1 + x)^{\frac{1}{6}}}\), we get
\(I = 6\left[ {\frac{{{{(x + 1)}^{\frac{1}{2}}}}}{3} – \frac{{{{(1 + x)}^{\frac{1}{3}}}}}{2} + {{(1 + x)}^{\frac{1}{6}}} – \ln \left( {{{(x + 1)}^{\frac{1}{6}}} + 1} \right)} \right] + C\)

Q.5. Evaluate \(\int {\sqrt {\left( {\frac{{x + 1}}{{x + 2}}} \right)} } \frac{{dx}}{{x + 3}}\).
Ans: Put \(\frac{{x + 1}}{{x + 2}} = {z^2}\)
\( \Rightarrow x = \frac{{2{z^2} – 1}}{{1 – {z^2}}} = – 2 + \frac{1}{{1 – {z^2}}}\) and \(x + 3 = \frac{{2 – {z^2}}}{{1 – {z^2}}}\)
\( \Rightarrow dx = 2zdz/{\left( {1 – {z^2}} \right)^2}\)
\(\therefore \,I = \int z \cdot \frac{{2zdz}}{{{{\left( {1 – {z^2}} \right)}^2}}}\left( {\frac{{1 – {z^2}}}{{2 – {z^2}}}} \right) = \int {\frac{{2{z^2}dz}}{{\left( {2 – {z^2}} \right)\left( {1 – {z^2}} \right)}}} \)
\( = \int {\left( {\frac{2}{{1 – {z^2}}} – \frac{4}{{2 – {z^2}}}} \right)} dz\)
\( = \ln \left| {\frac{{1 + z}}{{1 – z}}} \right| – \sqrt 2 \ln \left| {\frac{{\sqrt 2 + z}}{{\sqrt 2 – z}}} \right| + C\) where \(z = \sqrt {\left( {\frac{{x + 1}}{{x + 2}}} \right)} \)

Summary

A rational function combines two or more rational expressions into one. Writing a special type of sum of simpler fractions is called a partial fraction decomposition. Every individual term in the sum is called a partial fraction. There are four types of partial fractions. By using partial fractions, rational functions can be integrated.

Sometimes they can be avoided to integrate a rational function, and we can use different methods to integrate it. Irrational functions can be integrated by using suitable substitution for one of them using rationalisation. Certain types of integrals of irrational algebraic expressions can be reduced to integrals of a rational function by an appropriate change in the variables. Such transformation into a rational integral is called its rationalisation.

Frequently Asked Questions (FAQs)

Students might be having many questions with respect to the Integration of Rational and Irrational Functions. Here are a few commonly asked questions and answers.

Q.1. How do you integrate irrational functions?
Ans: We can integrate irrational function using suitable substitution or convert it into a rational function using rationalisation.

Q.2. What are rational and irrational functions?
Ans: A rational function is the ratio of polynomials to which the polynomial denominator should not be equal to zero. An irrational function is often considered to be a function that contains a radical sign.

Q.3. What is rational function integration?
Ans: Rational functions can be integrated by using several methods like using partial fractions and or using the formula of integration of special type of rational functions.

Q.4. Are rational functions integrable?
Ans: Yes, rational functions are integrated using suitable methods, and in substitution, we can integrate rational functions.

Q.5. How do you integrate rational expressions?
Ans: To integrate a proper rational function, we can apply the method of partial fractions.This method allows turning the integral of a complicated rational function into the sum of integrals of simpler functions.

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