Solution

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

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

Equating the coefficients of $x$ and constant term, we obtain

$A+B=1$

$2 A+B=0$

On solving, we obtain

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

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

$\Rightarrow \int \frac{x}{(x+1)(x+2)} d x=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} d x$

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

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

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

Solution

Let $\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}$

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

Equating the coefficients of $x$ and constant term, we obtain

$A+B=0$

$-3 A+3 B=1$

On solving, we obtain

$A=-\frac{1}{6}$ and $B=\frac{1}{6}$

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

$\Rightarrow \int \frac{1}{(x^{2}-9)} d x=\int(\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}) d x$

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

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

Solution

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

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

Substituting $x=1,2$, and 3 respectively in equation (1), we obtain

$A=1, B=-5$, and $C=4$

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

$\Rightarrow \int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}} d x$

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

Solution

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

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

Substituting $x=1,2$, and 3 respectively in equation (1), we obtain

$A=\frac{1}{2}, B=-2$, and $C=\frac{3}{2}$

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

$\Rightarrow \int \frac{x}{(x-1)(x-2)(x-3)} d x=\int{\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}} d x$

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

Solution

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

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

Substituting $x=-1$ and -2 in equation (1), we obtain

$A=-2$ and $B=4$ $\therefore \frac{2 x}{(x+1)(x+2)}=\frac{-2}{(x+1)}+\frac{4}{(x+2)}$

$\Rightarrow \int \frac{2 x}{(x+1)(x+2)} d x=\int{\frac{4}{(x+2)}-\frac{2}{(x+1)}} d x$

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

Solution

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $(1-x^{2})$ by $x(1-2 x)$, we obtain

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

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

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

Substituting $x=0$ and $\frac{1}{2}$ in equation (1), we obtain

$A=2$ and $B=3$

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

Substituting in equation (1), we obtain

$ \begin{aligned} & \frac{1-x^{2}}{x(1-2 x)}=\frac{1}{2}+\frac{1}{2}{\frac{2}{x}+\frac{3}{(1-2 x)}} \\ & \Rightarrow \int \frac{1-x^{2}}{x(1-2 x)} d x=\int{\frac{1}{2}+\frac{1}{2}(\frac{2}{x}+\frac{3}{1-2 x})} d x \\ & =\frac{x}{2}+\log |x|+\frac{3}{2(-2)} \log |1-2 x|+C \\ & =\frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C \end{aligned} $

Solution

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

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

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

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

$A+C=0$

$-A+B=1$

$-B+C=0$

On solving these equations, we obtain

$A=-\frac{1}{2}, B=\frac{1}{2}$, and $C=\frac{1}{2}$

From equation (1), we obtain

$ \begin{aligned} & \therefore \frac{x}{(x^{2}+1)(x-1)}=\frac{(-\frac{1}{2} x+\frac{1}{2})}{x^{2}+1}+\frac{\frac{1}{2}}{(x-1)} \\ & \Rightarrow \int \frac{x}{(x^{2}+1)(x-1)}=-\frac{1}{2} \int \frac{x}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x^{2}+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x \\ & \quad=-\frac{1}{4} \int \frac{2 x}{x^{2}+1} d x+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C \end{aligned} $

Consider $\int \frac{2 x}{x^{2}+1} d x$, let $(x^{2}+1)=t \Rightarrow 2 x d x=d t$

$\Rightarrow \int \frac{2 x}{x^{2}+1} d x=\int \frac{d t}{t}=\log |t|=\log |x^{2}+1|$

$ \begin{aligned} \therefore \int \frac{x}{(x^{2}+1)(x-1)} & =-\frac{1}{4} \log |x^{2}+1|+\frac{1}{2} \tan ^{-1} x+\frac{1}{2} \log |x-1|+C \\ & =\frac{1}{2} \log |x-1|-\frac{1}{4} \log |x^{2}+1|+\frac{1}{2} \tan ^{-1} x+C \end{aligned} $

Solution

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

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

Substituting $x=1$, we obtain

$B=\frac{1}{3}$

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

$A+C=0$

$-2 A+2 B+C=0$

On solving, we obtain

$ \begin{aligned} & A=\frac{2}{9} \text{ and } C=\frac{-2}{9} \\ & \begin{aligned} & \therefore \frac{x}{(x-1)^{2}(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1)^{2}}-\frac{2}{9(x+2)} \\ & \Rightarrow \int \frac{x}{(x-1)^{2}(x+2)} d x=\frac{2}{9} \int \frac{1}{(x-1)} d x+\frac{1}{3} \int \frac{1}{(x-1)^{2}} d x-\frac{2}{9} \int \frac{1}{(x+2)} d x \\ &=\frac{2}{9} \log |x-1|+\frac{1}{3}(\frac{-1}{x-1})-\frac{2}{9} \log |x+2|+C \\ &=\frac{2}{9} \log |\frac{x-1}{x+2}|-\frac{1}{3(x-1)}+C \end{aligned} \end{aligned} $

Solution

$\frac{3 x+5}{x^{3}-x^{2}-x+1}=\frac{3 x+5}{(x-1)^{2}(x+1)}$

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

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

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

Substituting $x=1$ in equation (1), we obtain

$B=4$

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

$A+C=0$

$B-2 C=3$

On solving, we obtain

$A=-\frac{1}{2}$ and $C=\frac{1}{2}$ $\therefore \frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)}$

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

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

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

Solution

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

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

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

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

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

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

$B=-\frac{1}{10}, A=\frac{5}{2}$, and $C=-\frac{24}{5}$ $\therefore \frac{2 x-3}{(x+1)(x-1)(2 x+3)}=\frac{5}{2(x+1)}-\frac{1}{10(x-1)}-\frac{24}{5(2 x+3)}$

$\Rightarrow \int \frac{2 x-3}{(x^{2}-1)(2 x+3)} d x=\frac{5}{2} \int \frac{1}{(x+1)} d x-\frac{1}{10} \int \frac{1}{x-1} d x-\frac{24}{5} \int \frac{1}{(2 x+3)} d x$

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

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

Solution

$\frac{5 x}{(x+1)(x^{2}-4)}=\frac{5 x}{(x+1)(x+2)(x-2)}$

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

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

Substituting $x=-1,-2$, and 2 respectively in equation (1), we obtain

$ \begin{aligned} & A=\frac{5}{3}, B=-\frac{5}{2}, \text{ and } C=\frac{5}{6} \\ & \begin{aligned} \therefore \frac{5 x}{(x+1)(x+2)(x-2)} & =\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)} \\ \Rightarrow \int \frac{5 x}{(x+1)(x^{2}-4)} d x & =\frac{5}{3} \int \frac{1}{(x+1)} d x-\frac{5}{2} \int \frac{1}{(x+2)} d x+\frac{5}{6} \int \frac{1}{(x-2)} d x \\ & =\frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+C \end{aligned} \end{aligned} $

Solution

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing $(x^{3}+x+1)$ by $x^{2}-1$, we obtain

$\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2 x+1}{x^{2}-1}$

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

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

Substituting $x=1$ and -1 in equation (1), we obtain

$ A=\frac{1}{2} \text{ and } B=\frac{3}{2} $

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

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

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

Solution

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

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

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

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

$A-B=0$

$B-C=0$

$A+C=2$

On solving these equations, we obtain

$A=1, B=1$, and $C=1$

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

$\Rightarrow \int \frac{2}{(1-x)(1+x^{2})} d x=\int \frac{1}{1-x} d x+\int \frac{x}{1+x^{2}} d x+\int \frac{1}{1+x^{2}} d x$

$=-\int \frac{1}{x-1} d x+\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x+\int \frac{1}{1+x^{2}} d x$

$=-\log |x-1|+\frac{1}{2} \log |1+x^{2}|+\tan ^{-1} x+C$

Solution

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

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

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

$2 A+B=-1 \Rightarrow B=-7$ $\therefore \frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}$

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

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

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

Solution

$\frac{1}{(x^{4}-1)}=\frac{1}{(x^{2}-1)(x^{2}+1)}=\frac{1}{(x+1)(x-1)(1+x^{2})}$

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

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

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

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

Equating the coefficient of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+B+C=0$

$-A+B+D=0$

$A+B-C=0$

$-A+B-D=1$

On solving these equations, we obtain

$A=-\frac{1}{4}, B=\frac{1}{4}, C=0$, and $D=-\frac{1}{2}$ $\therefore \frac{1}{x^{4}-1}=\frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^{2}+1)}$

$\Rightarrow \int \frac{1}{x^{4}-1} d x=-\frac{1}{4} \log |x-1|+\frac{1}{4} \log |x-1|-\frac{1}{2} \tan ^{-1} x+C$

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

Solution

$\frac{1}{x(x^{n}+1)}$

Multiplying numerator and denominator by $x^{n-1}$, we obtain

$\frac{1}{x(x^{n}+1)}=\frac{x^{n-1}}{x^{n-1} x(x^{n}+1)}=\frac{x^{n-1}}{x^{n}(x^{n}+1)}$

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

$\therefore \int \frac{1}{x(x^{n}+1)} d x=\int \frac{x^{n-1}}{x^{n}(x^{n}+1)} d x=\frac{1}{n} \int \frac{1}{t(t+1)} d t$

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

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

Substituting $t=0,-1$ in equation (1), we obtain

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

$\therefore \frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{(1+t)}$ $\Rightarrow \int \frac{1}{x(x^{n}+1)} d x=\frac{1}{n} \int{\frac{1}{t}-\frac{1}{(t+1)}} d x$

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

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

$=\frac{1}{n} \log |\frac{x^{n}}{x^{n}+1}|+C$

Solution

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

Let $\sin x=t \Rightarrow \cos x d x=d t$

$\therefore \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \frac{d t}{(1-t)(2-t)}$

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

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

Substituting $t=2$ and then $t=1$ in equation (1), we obtain

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

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

$ \begin{aligned} \Rightarrow \int \frac{\cos x}{(1-\sin x)(2-\sin x)} d x & =\int{\frac{1}{1-t}-\frac{1}{(2-t)}} d t \\ & =-\log |1-t|+\log |2-t|+C \\ & =\log |\frac{2-t}{1-t}|+C \\ & =\log |\frac{2-\sin x}{1-\sin x}|+C \end{aligned} $

Solution

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

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

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

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

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

Equating the coefficients of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+C=0$

$B+D=4$

$4 A+3 C=0$

$4 B+3 D=10$

On solving these equations, we obtain

$A=0, B=-2, C=0$, and $D=6$

$\therefore \frac{4 x^{2}+10}{(x^{2}+3)(x^{2}+4)}=\frac{-2}{(x^{2}+3)}+\frac{6}{(x^{2}+4)}$ $\frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)}=1-(\frac{-2}{(x^{2}+3)}+\frac{6}{(x^{2}+4)})$

$\Rightarrow \int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} d x=\int{1+\frac{2}{(x^{2}+3)}-\frac{6}{(x^{2}+4)}} d x$

$={1+\frac{2}{x^{2}+(\sqrt{3})^{2}}-\frac{6}{x^{2}+2^{2}}}$

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

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

Solution

$\frac{2 x}{(x^{2}+1)(x^{2}+3)}$

Let $x^{2}=t \Rightarrow 2 x d x=d t$

$\therefore \int \frac{2 x}{(x^{2}+1)(x^{2}+3)} d x=\int \frac{d t}{(t+1)(t+3)}$

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

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

Substituting $t=-3$ and $t=-1$ in equation (1), we obtain $A=\frac{1}{2}$ and $B=-\frac{1}{2}$

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

$\Rightarrow \int \frac{2 x}{(x^{2}+1)(x^{2}+3)} d x=\int{\frac{1}{2(t+1)}-\frac{1}{2(t+3)}} d t$

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

$=\frac{1}{2} \log |\frac{t+1}{t+3}|+C$

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

Solution

$\frac{1}{x(x^{4}-1)}$

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

$ \begin{aligned} & \frac{1}{x(x^{4}-1)}=\frac{x^{3}}{x^{4}(x^{4}-1)} \\ & \therefore \int \frac{1}{x(x^{4}-1)} d x=\int \frac{x^{3}}{x^{4}(x^{4}-1)} d x \end{aligned} $

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

$\therefore \int \frac{1}{x(x^{4}-1)} d x=\frac{1}{4} \int \frac{d t}{t(t-1)}$

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

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

Substituting $t=0$ and 1 in (1), we obtain

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

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

$\Rightarrow \int \frac{1}{x(x^{4}-1)} d x=\frac{1}{4} \int{\frac{-1}{t}+\frac{1}{t-1}} d t$

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

$=\frac{1}{4} \log |\frac{t-1}{t}|+C$

$=\frac{1}{4} \log |\frac{x^{4}-1}{x^{4}}|+C$

Solution

$\frac{1}{(e^{x}-1)}$

Let $e^{x}=t \Rightarrow e^{x} d x=d t$

$\Rightarrow \int \frac{1}{e^{x}-1} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t$

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

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

Substituting $t=1$ and $t=0$ in equation (1), we obtain

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

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

$\Rightarrow \int \frac{1}{t(t-1)} d t=\log |\frac{t-1}{t}|+C$

$ =\log |\frac{e^{x}-1}{e^{x}}|+C $

Solution

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

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

Substituting $x=1$ and 2 in (1), we obtain

$A=-1$ and $B=2$ $\therefore \frac{x}{(x-1)(x-2)}=-\frac{1}{(x-1)}+\frac{2}{(x-2)}$

$\Rightarrow \int \frac{x}{(x-1)(x-2)} d x=\int{\frac{-1}{(x-1)}+\frac{2}{(x-2)}} d x$

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

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

Hence, the correct Answer is B.

Solution

Let $\frac{1}{x(x^{2}+1)}=\frac{A}{x}+\frac{B x+C}{x^{2}+1}$

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

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 $C=0$ $\therefore \frac{1}{x(x^{2}+1)}=\frac{1}{x}+\frac{-x}{x^{2}+1}$

$\Rightarrow \int \frac{1}{x(x^{2}+1)} d x=\int{\frac{1}{x}-\frac{x}{x^{2}+1}} d x$

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

Hence, the correct Answer is A.