Solution

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 $x$ and constant term, we obtain

$A+B=1$

$2 A+B=0$

On solving, we obtain

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

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

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

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

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

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

Solution

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

$A+B=0$

$-3 A+3 B=1$

On solving, we obtain

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

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

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

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

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

Solution

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

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

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

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

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

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

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

Solution

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

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

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

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

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

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

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

Solution

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

$2 x=A(x+2)+B(x+1) \qquad…(1)$

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

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

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

$ =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

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

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

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

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

$A=2$ and $B=3$

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

Substituting in equation (1), we obtain

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

Solution

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

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

$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

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

From equation (1), we obtain

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

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

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

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

Solution

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

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

Substituting $x=1$, we obtain

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

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

Solution

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

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

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

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

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

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

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

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

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

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

Solution

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

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

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

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

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

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

$B=-\dfrac{1}{10}, A=\dfrac{5}{2}$, and $C=-\dfrac{24}{5}$

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

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

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

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

Solution

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

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

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

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

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

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

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

$2 x+1=A(x-1)+B(x+1) \qquad…(1)$

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

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

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

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

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

Solution

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

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

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

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 $C=1$

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

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

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

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

Solution

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

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

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

$2 A+B=-1 \Rightarrow B=-7$

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

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

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

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

Solution

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

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

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

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

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

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

$A=-\dfrac{1}{4}, B=\dfrac{1}{4}, C=0$, and $D=-\dfrac{1}{2}$

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

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

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

Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution

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

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

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

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

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

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

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

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

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

Solution

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

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

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

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

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

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 $D=6$

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

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

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

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

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

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

Solution

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

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

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

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

$1=A(t+3)+B(t+1) \qquad …(1)$

Substituting $t=-3$ and $t=-1$ in equation (1), we obtain

$A=\dfrac{1}{2}$ and $B=-\dfrac{1}{2}$

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

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

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

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

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

Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution

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

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

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

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

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

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

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

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

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

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

Solution

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

$x=A(x-2)+B(x-1) \qquad…(1)$

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

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

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

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

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

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

Hence, the correct Answer is B.

Solution

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

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

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 $C=0$

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

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

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

Hence, the correct Answer is A.