Chapter 7: Integrals

Ans :

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

$ \Rightarrow \int {\dfrac{1}{{\left( {{{\text{e}}^x} - 1} \right)}}} {\text{dx}} = \int {\dfrac{1}{{{\text{t}} - 1}}}  \times \dfrac{{{\text{dt}}}}{{\text{t}}} - \int {\dfrac{1}{{{\text{t}}({\text{t}} - 1)}}} {\text{dt}}$

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 ${\text{t}}$ and constant, we obtain

$A + B = 0$ and $ - A = 1$

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

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

$ \Rightarrow \int {\dfrac{1}{{t(t - 1)}}} dt = \log \left| {\dfrac{{t - 1}}{t}} \right| + C$

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

Where $C$ is an arbitrary constant.

Ans:

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

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

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

$A--1$ and $B = 2$

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

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

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

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

Hence, the correct Answer is $B$.

Ans :

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

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

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

$c = 0$

$A = 1$

On solving these equations, we obtain $A = 1,B =  - 1$, and $C = 0$

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

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

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

Hence, the correct Answer is ${\text{A}}$. 

Where $C$ is an arbitrary constant.

Alter:

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

Let ${x^2} = t$, therefore, $2xdx = dt$ $\therefore \int {\dfrac{x}{{{x^2}\left( {{x^2} + 1} \right)}}} dx = \dfrac{1}{2}\int {\dfrac{{dt}}{{t(t + 1)}}}  = \dfrac{1}{2}\int {\dfrac{{(t + 1) - t}}{{t(t + 1)}}} dt = \dfrac{1}{2}\int \frac{1}{t} -  \dfrac{1}{{t + 1}}dt$

$ - \dfrac{1}{2}[\log t - \log (t + 1)] + C$

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

Where $C$ is an arbitrary constant.