Chapter 7: Integrals

Ans: Let $\dfrac{x}{{(x + 1)(x + 2)}} = \dfrac{A}{{(x + 1)}} + \dfrac{B}{{(x + 2)}}$

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

Equating the coefficients of ${\text{x}}$ and constant term, we obtain $A + B = 1$

$2A + B = 0$

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

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

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

$ =  - \log |x + 1| + 2\log |x + 2| + C$

$ = \log {(x + 2)^2} - \log |x + 1| + C$

$ = \log \dfrac{{{{(x + 2)}^2}}}{{(x + 1)}} + C$

Where $C$ is an arbitrary constant

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

$1 = A(x - 3) + B(x + 3)$

Equating the coefficients of $x$ and constant term, we obtain $A + B = 0$

$ - 3A + 3B = 1$

On solving. we obtain

${\text{A}} =  - \dfrac{1}{6}$ and ${\text{B}} = \dfrac{1}{6}$

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

$ \Rightarrow \int {\dfrac{1}{{\left( {{x^2} - 9} \right)}}} dx = \int {\left( {\dfrac{{ - 1}}{{6(x + 3)}} + \dfrac{1}{{6(x - 3)}}} \right)} dx$

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

$ = \dfrac{1}{6}\log \dfrac{{|(x - 3)|}}{{|(x + 3)|}} + C$

Where $C$ is an arbitrary constant

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

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

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

$ - 5A - 4B - 3C = 3$

$6A + 3B + 2C =  - 1$

Solving these equations, we obtain ${\text{A}} = 1,\;{\text{B}} =  - 5$, and $C = 4$

$\dfrac{{3x - 1}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{1}{{(x - 1)}} - \dfrac{5}{{(x - 2)}} + \dfrac{4}{{(x - 3)}}$

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

$ = \log |x - 1| - 5\log |x - 2| + 4\log |x - 3| + C$

Where $C$ is an arbitrary constant.

Ans: $\dfrac{x}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{(x - 2)}} + \dfrac{C}{{(x - 3)}}$

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

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

$45 - 3C = 1$

$6A + 4B + 2C = 0$

Solving these equations, we obtain $A = \dfrac{1}{2},B = 2$ and $C = \dfrac{3}{2}$

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

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

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

Where $C$ is an arbitrary constant.

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

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

$ \ldots (1)$

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

$2A + B = 0$

Solving these equations, we obtain $A =  - 2$ and ${\mathbf{B}} = 4$

$\therefore \dfrac{{2x}}{{(x + 1)(x + 2)}} = \dfrac{{ - 2}}{{(x + 1)}} + \dfrac{4}{{(x + 2)}}$

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

$ = 4\log |x + 2| - 2\log |x + 1| + C$

Where $C$ is an arbitrary constant.

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

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

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

Equating the coefficients of ${x^2},x$ and constant term, we obtain $ - 2A + B =  - 1$

And $A = 2$ Solving these equations, we obtain $A = 2$ and $B = 3$ $\therefore \dfrac{{2 - x}}{{x(1 - 2x)}} = \dfrac{2}{x} + \dfrac{3}{{1 - 2x}}$

Substituting in equation (1), we obtain $\dfrac{{1 - {x^2}}}{{x(1 - 2x)}} = \dfrac{1}{2} + \dfrac{1}{2}\left\{ {\dfrac{2}{x} + \dfrac{3}{{(1 - 2x)}}} \right\}$

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

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

$ = \dfrac{x}{2} + \log |x| - \dfrac{3}{4}\log |1 - 2x| + c$

Where $C$ is an arbitrary constant.

Ans: Let $\dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{Ax + B}}{{\left( {{x^2} + 1} \right)}} + \dfrac{C}{{(x - 1)}}$

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

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

Equating the coefficients of ${x^2},x$, and constant term, we obtain

A $ - A + B = 1$

$ - B + C = 0$

On solving these equations, we obtain ${\text{A}} =  - \dfrac{1}{2},\;{\text{B}} = \dfrac{1}{2}$, and ${\text{C}} = \dfrac{1}{2}$

From equation (1), vre obtain $\therefore \dfrac{x}{{\left( {{x^2} + 1} \right)(x - 1)}} = \dfrac{{\left( { - \dfrac{1}{2}x + \dfrac{1}{2}} \right)}}{{{x^2} + 1}} + \dfrac{{\dfrac{1}{2}}}{{(x - 1)}}$

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

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

Consider $\int {\dfrac{{2x}}{{{x^2} + 1}}} dx$, let $\left( {{x^2} + 1} \right) = t \Rightarrow 2xdx = dt$

$ \Rightarrow \int {\dfrac{{2x}}{{{x^2} + 1}}} dx - \int {\dfrac{{dt}}{t}}  - \log |t| - \log \left| {{x^2} + 1} \right|$

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

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

Where $C$ is an arbitrary constant.

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

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

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

$A + B - 2C = 1$

$ - 2\;{\text{A}} + 2\;{\text{B}} + {\text{C}} = 0$

On solving. we obtain $A = \dfrac{2}{9}$ and $C = \dfrac{{ - 2}}{9}$

${\text{B}} = \dfrac{1}{3}$

$\therefore \dfrac{x}{{{{(x - 1)}^2}(x + 2)}} = \dfrac{2}{{9(x - 1)}} + \dfrac{1}{{3{{(x - 1)}^2}}} - \dfrac{2}{{9(x + 2)}}$

$ \Rightarrow \int {\dfrac{x}{{{{(x - 1)}^2}(x + 2)}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x - 1)}}} dx + \dfrac{1}{3}\int {\dfrac{1}{{{{(x - 1)}^2}}}} dx - \dfrac{2}{9}\int {\dfrac{1}{{(x + 2)}}} dx$

$ = \dfrac{2}{9}\log |x - 1| + \dfrac{1}{3}\left( {\dfrac{{ - 1}}{{x - 1}}} \right) - \dfrac{2}{9}\log |x + 2| + C$

$ = \dfrac{2}{9}\log \left| {\dfrac{{x - 1}}{{x + 2}}} \right| - \dfrac{1}{{3(x - 1)}} + C$

Where $C$ is an arbitrary constant.

Ans: $\dfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}} = \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}}$

Let $\dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{A}{{(x - 1)}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{(x + 1)}}$

$3x + 5 = A(x - 1)(x + 1) + B(x + 1) + C{(x - 1)^2}$

$3x + 5 = A\left( {{x^2} - 1} \right) + B(x + 1) + C\left( {{x^2} + 1 - 2x} \right)$

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

$B - 2C - 3$

$ - A + B + C = 5$

On solving. we obtain $B = 4$

${\text{A}} =  - \dfrac{1}{2}$ and ${\text{C}} = \dfrac{1}{2}$

$\therefore \dfrac{{3x + 5}}{{{{(x - 1)}^2}(x + 1)}} = \dfrac{{ - 1}}{{2(x - 1)}} + \dfrac{4}{{{{(x - 1)}^2}}} + \dfrac{1}{{2(x + 1)}}$

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

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

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

Where $C$ is an arbitrary constant.

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

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

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

$ \Rightarrow (2x - 3) = A\left( {2{x^2} + x - 3} \right) + B\left( {2{x^2} + 5x + 3} \right) + C\left( {{x^2} - 1} \right)$

$ \Rightarrow (2x - 3) = (2A + 2B + C){x^2} + (A + 5B)x + ( - 3A + 3B - C)$

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

$A + 5B = 2$

$ - 3A + 3B - C =  - 3$

On solving, we obtain ${\text{B}} =  - \dfrac{1}{{10}},\;{\text{A}} = \dfrac{5}{2}$, and ${\text{C}} =  - \dfrac{{24}}{5}$

$\therefore \dfrac{{2x - 3}}{{(x + 1)(x - 1)(2x + 3)}} = \dfrac{5}{{2(x + 1)}} - \dfrac{1}{{10(x - 1)}} - \dfrac{{24}}{{5(2x + 3)}}$

$ \Rightarrow \int {\dfrac{{2x - 3}}{{\left( {{x^2} - 1} \right)(2x + 3)}}} dx = \dfrac{5}{2}\int {\dfrac{1}{{(x + 1)}}} dx - \dfrac{1}{{10}}\int {\dfrac{1}{{x - 1}}} dx - \dfrac{{24}}{5}\int {\dfrac{1}{{(2x + 3)}}} dx$

$ = \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{24}}{{5 \times 2}}\log |2x + 3|$

$ = \dfrac{5}{2}\log |x + 1| - \dfrac{1}{{10}}\log |x - 1| - \dfrac{{12}}{5}\log |2x + 3| + C$

Where $C$ is an arbitrary constant.