STD 12 - 7.1 inderfinite integral — JEE कक्षा 11 साइन्स — Question

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

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 $C=0$

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

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

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

Hence, the correct Answer is $A$.

Alter:

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

Let $x^{2}=t,$ therefore, $2 x\, d x=d t$

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

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

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