Chapter 7: Integrals

Ans : $\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}} = \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}}$

Let $\dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}} + \dfrac{C}{{(x - 2)}}$

$5x = A(x + 2)(x - 2) + B(x + 1)(x - 2) + C(x + 1)(x + 2)$

Equating the coefficients of ${x^2},x$ and constant, we obtain $A + B + C = 0$

$B + 3C = 5$ and $4A - 2B + 2C = 0$

On solving. we obtain $A - \dfrac{5}{3},B -  - \dfrac{5}{2}$, and $C - \dfrac{5}{6}$

$\therefore \dfrac{{5x}}{{(x + 1)(x + 2)(x - 2)}} = \dfrac{5}{{3(x + 1)}} +  - \dfrac{5}{{2(x + 2)}} + \dfrac{5}{{6(x - 2)}}$

$ \Rightarrow \int {\dfrac{{5x}}{{(x + 1)\left( {{x^2} - 4} \right)}}} dx = \dfrac{5}{3}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{5}{2}\int {\dfrac{1}{{(x + 2)}}} dx + \dfrac{5}{6}\int {\dfrac{1}{{(x - 2)}}} dx$

$ = \dfrac{5}{3}\log |x + 1| - \dfrac{5}{2}\log |x + 2| + \dfrac{5}{6}\log |x - 2| + C$

Where $C$ is an arbitrary constant.

Ans: It can be seen that the given integrand is not a proper fraction. Therefore, on dividing $\left( {{x^3} + x + 1} \right)$ by ${x^2} - 1$, we obtain $\dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{{2x + 1}}{{{x^2} - 1}}$

Let $\dfrac{{2x + 1}}{{{x^2} - 1}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}}$

$2x + 1 = A(x - 1) + B(x + 1)$

Equating the coefficients of $x$ and constant, we obtain $A + B = 2$

$ - A + B = 1$

On solving. we obtain $A - \dfrac{1}{2}$ and $B - \dfrac{3}{2}$

$\therefore \dfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \dfrac{1}{{2(x + 1)}} + \dfrac{3}{{2(x - 1)}}$

$ \Rightarrow \int {\dfrac{{{x^3} + x + 1}}{{{x^2} + 1}}} dx = \int x dx + \dfrac{1}{2}\int {\dfrac{1}{{(x + 1)}}} dx + \dfrac{3}{2}\int {\dfrac{1}{{(x - 1)}}} dx$

$ = \dfrac{{{x^2}}}{2} + \log |x + 1| + \dfrac{3}{2}\log |x - 1| + C$

Where $C$ is an arbitrary constant.

Ans:

Let $\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(1 - x)}} + \dfrac{{Bx + C}}{{\left( {1 + {x^2}} \right)}}$

$2 = A\left( {1 + {x^2}} \right) + (Bx + C)(1 - x)$

$2 = A + A{x^2} + Bx - B{x^2} + C - Cx$

Equating the coefficient of ${x^2},x$, and constant term, we obtain ${\text{A}} - {\text{B}} = 0$

${\mathbf{B}} - {\mathbf{C}} = {\mathbf{0}}$

$A + C = 2$

On solving these equations, we obtain $A = 1,B = 1$, and $C = 1$

$\therefore \dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}} = \dfrac{1}{{1 - x}} + \dfrac{{x + 1}}{{1 + {x^2}}}$

$ \Rightarrow \int {\dfrac{2}{{(1 - x)\left( {1 + {x^2}} \right)}}} dx = \int {\dfrac{1}{{1 - x}}} dx + \int {\dfrac{x}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$ =  - \int {\dfrac{1}{{1 - x}}} dx + \dfrac{1}{2}\int {\dfrac{{2x}}{{1 + {x^2}}}} dx + \int {\dfrac{1}{{1 + {x^2}}}} dx$

$ =  - \log |x - 1| + \dfrac{1}{2}\log \left| {1 + {x^2}} \right| + {\tan ^{ - 1}}x + C$

Where $C$ is an arbitrary constant.

Ans:

Let $\dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{A}{{(x + 2)}} + \dfrac{B}{{{{(x + 2)}^2}}}$

$ \Rightarrow 3x - 1 = A(x + 2) + B$

Equating the coefficient of $x$ and constant term, we obtain $A = 3$

$2A + B =  - 1 \Rightarrow \mathbb{B} -  - 7$

$\therefore \dfrac{{3x - 1}}{{{{(x + 2)}^2}}} = \dfrac{3}{{(x + 2)}} - \dfrac{7}{{{{(x + 2)}^2}}}$

$ \Rightarrow \int {\dfrac{{3x - 1}}{{{{(x + 2)}^2}}}} dx = 3\int {\dfrac{1}{{(x + 2)}}} dx - 7\int {\dfrac{1}{{{{(x + 2)}^2}}}} dx$

$ = 3\log |x + 2| - 7\left( {\dfrac{{ - 1}}{{(x + 2)}}} \right) + C$

$ = 3\log |x + 2| + \dfrac{7}{{(x + 2)}} + C$

Where $C$ is an arbitrary constant.

Ans :

$\dfrac{1}{{\left( {{x^4} - 1} \right)}} - \dfrac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}} - \dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}}$

Let $\dfrac{1}{{(x + 1)(x - 1)\left( {1 + {x^2}} \right)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x - 1)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 1} \right)}}$

$1 = A(x - 1)\left( {1 + {x^2}} \right) + B(x + 1)\left( {1 + {x^2}} \right) + (Cx + D)\left( {{x^2} - 1} \right)$

$1 = A\left( {{x^3} + x - {x^2} - 1} \right) + B\left( {{x^3} + x + {x^2} + 1} \right) + C{x^3} + D{x^2} - Cx - D$

$1 = (A + B + C){x^3} + ( - A + B + D){x^2} + (A + B - C)x + ( - A + B - D)$

Equating the coefficient of ${x^3},{x^2},x$, and constant term, we obtain $A + B + C = 0$

$ - A + B + D = 0$

$A + B - C = 0$

$ - A + B - D = 1$

${\text{A}} =  - \dfrac{1}{4},\;{\text{B}} = \dfrac{1}{4},{\text{C}} = {\text{O}}$, and ${\text{D}} =  - \dfrac{1}{2}$

$\therefore \dfrac{1}{{\left( {{x^n} - 1} \right)}} = \dfrac{{ - 1}}{{4(x + 1)}} + \dfrac{1}{{4(x - 1)}} + \dfrac{1}{{2\left( {{x^2} + 1} \right)}}$

$ \Rightarrow \int {\dfrac{1}{{{x^4} - 1}}} dx -  - \dfrac{1}{4}\log |x - 1| + \dfrac{1}{4}\log |x - 1| - \dfrac{1}{2}{\tan ^1}x + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{x - 1}}{{x + 1}}} \right| - \dfrac{1}{2}{\tan ^1}x + C$

Where $C$ is an arbitrary constant.

Ans:

$\dfrac{1}{{x\left( {{x^n} + 1} \right)}}$

Multiplying numerator and denominator by ${x^{n - 1}}$, we obtain $\dfrac{1}{{x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^{n - 1}}x\left( {{x^n} + 1} \right)}} = \dfrac{{{x^{n - 1}}}}{{{x^n}\left( {{x^n} + 1} \right)}}$

Let ${x^n} = t \Rightarrow n{x^{ - 1}}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^n} + 1} \right)}}} dx = \int {\dfrac{{{x^{n - 1}}}}{{{x^7}\left( {{x^n} + 1} \right)}}} dx = \dfrac{1}{n}\int {\dfrac{1}{{t(t + 1)}}} dt$

Let $\dfrac{1}{{t(t + 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t + 1)}}$

$1 = A(1 + t) + Bt$

Equating the coefficients of $t$ and constant, ve obtain $A = 1$ and $B =  - 1$

$\dfrac{1}{{t(t + 1)}} = \dfrac{1}{t} - \dfrac{1}{{(1 + t)}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^4} + 1} \right)}}} dx = \dfrac{1}{n}\int {\left\{ {\dfrac{1}{t} - \dfrac{1}{{(1 + t)}}} \right\}} dx$

$\dfrac{1}{n}[\log |t| - \log |t + 1|] + C$

$ = \dfrac{1}{n}\left[ {\log \left| {{x^n}} \right| - \log \left| {{x^n} + 1} \right|} \right] + C$

$ = \dfrac{1}{n}\log \left| {\dfrac{{{x^n}}}{{{x^2} + 1}}} \right| + C$

Where $C$ is an arbitrary constant.

Ans:

$\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}$

let $\sin x = t \Rightarrow \cos xdx = dt$

$\therefore \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\dfrac{{dt}}{{(1 - t)(2 - t)}}} $

Let $\dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{A}{{(1 - t)}} + \dfrac{B}{{(2 - t)}}$

$1 = A(2 - t) + B(1 - t)$

Equating the coefficients of $t$ and constant, wre obtain $ - A - B = 0$ and $2A + B = 1$

On solving. we obtain $A = 1$ and $B =  - 1$

$\therefore \dfrac{1}{{(1 - t)(2 - t)}} = \dfrac{1}{{(1 - t)}} - \dfrac{1}{{(2 - t)}}$

$ \Rightarrow \int {\dfrac{{\cos x}}{{(1 - \sin x)(2 - \sin x)}}} dx = \int {\left\{ {\dfrac{1}{{1 - t}} - \dfrac{1}{{(2 - t)}}} \right\}} dt$

$ = \log \left| {\dfrac{{2 - t}}{{1 - t}}} \right| + C$

$ = \log \left| {\dfrac{{2 - \sin x}}{{1 - \sin x}}} \right| + C$

Where $C$ is an arbitrary constant.

Ans:

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \dfrac{{4{x^2} + 10}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}$

Let $\dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 3} \right)}} + \dfrac{{Cx + D}}{{\left( {{x^2} + 4} \right)}}$

$4{x^2} + 10 = (Ax + B)\left( {{x^2} + 4} \right) + (Cx + D)\left( {{x^2} + 3} \right)$

$4{x^2} + 10 = A{x^3} + 4Ax + B{x^2} + 4B + C{x^3} + 3Cx + D{x^2} + 3D$

$4{x^2} + 10 = (A + C){x^2} + (B + D){x^2} + (4A + 3C)x + (4B + 3D)$

Equating the coefficients of ${x^3},{x^2},x$ and constant term, we obtain $A + C = 0$

$B + D = 4$

$4A + 3C = 0$

$4B + 3D = 10$

On solving these equations, we obtain ${\text{A}} = 0,\;{\text{B}} =  - 2,{\text{C}} = 0$, and ${\text{D}} = 6$

$\therefore \dfrac{{\left( {4{x^2} + 10} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}$

$\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \left( {\dfrac{{ - 2}}{{\left( {{x^2} + 3} \right)}} + \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right)$

$ \Rightarrow \int {\dfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}}} dx = \int {\left\{ {1 + \dfrac{2}{{\left( {{x^2} + 3} \right)}} - \dfrac{6}{{\left( {{x^2} + 4} \right)}}} \right\}} dx$

$ = \int {\left\{ {1 + \dfrac{2}{{{x^2} + {{(\sqrt 3 )}^2}}} - \dfrac{6}{{{x^2} + {2^2}}}} \right\}} $

$ = x + 2\left( {\dfrac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\dfrac{x}{{\sqrt 3 }}} \right) - 6\left( {\dfrac{1}{2}{{\tan }^{ - 1}}\dfrac{x}{2}} \right) + C$

$ = x + \dfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{x}{{\sqrt 3 }} - 3{\tan ^{ - 1}}\dfrac{x}{2} + C$

Where $C$ is an arbitrary constant.

Ans:

$\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}$

Let ${x^2} = t \Rightarrow 2xdx = dt$

$\therefore \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\dfrac{{dt}}{{(t + 1)(t + 3)}}} $

Let $\dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{A}{{(t + 1)}} + \dfrac{B}{{(t + 3)}}$

$1 = A(t + 3) + B(t + 1)$

Equating the coefficients of ${\text{t}}$ and constant, we obtain $A + B = 0$ and $3A + B = 1$

On solving. we obtain ${\text{A}} = \dfrac{1}{2}$ and ${\text{B}} =  - \dfrac{1}{2}$

$\therefore \dfrac{1}{{(t + 1)(t + 3)}} = \dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}$

$ \Rightarrow \int {\dfrac{{2x}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}} dx = \int {\left\{ {\dfrac{1}{{2(t + 1)}} - \dfrac{1}{{2(t + 3)}}} \right\}} dt$

$ = \dfrac{1}{2}\log |(t + 1)| - \dfrac{1}{2}\log |t + 3| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{t + 1}}{{t + 3}}} \right| + C$

$ = \dfrac{1}{2}\log \left| {\dfrac{{{x^2} + 1}}{{{x^2} + 3}}} \right| + C$

Where $C$ is an arbitrary constant.

Ans:

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}}$

Multiplying numerator and denominator by ${x^3}$, we obtain

$\dfrac{1}{{x\left( {{x^4} - 1} \right)}} = \dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \int {\dfrac{{{x^3}}}{{{x^4}\left( {{x^4} - 1} \right)}}} dx$

Let ${x^4} = t \Rightarrow 4{x^3}dx = dt$

$\therefore \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\dfrac{{dt}}{{t(t - 1)}}} $

Let $\dfrac{1}{{t(t - 1)}} = \dfrac{A}{t} + \dfrac{B}{{(t - 1)}}$

$1 = {\text{A}}({\text{t}} - 1) + {\text{Bt}}$

Equating the coefficients of $t$ and constant, we obtain $A + B = 0$ and $ - A = 1$

${\text{A}} =  - 1$ and ${\mathbf{B}} = 1$

$ \Rightarrow \dfrac{1}{{t(t - 1)}} = \dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}$

$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \dfrac{1}{4}\int {\left\{ {\dfrac{{ - 1}}{t} + \dfrac{1}{{t - 1}}} \right\}} dt$

$ - \dfrac{1}{4}[ - \log |t| + \log |t - 1|] + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{t - 1}}{t}} \right| + C$

$ = \dfrac{1}{4}\log \left| {\dfrac{{{x^4} - 1\mid }}{{{x^4}}}} \right| + C$

Where $C$ is an arbitrary constant.