Question

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

• A. $\ln \frac{\left|x\right|}{\sqrt{x^2+1}}+\frac{1}{2(x^2+1)}+K$
• B. $\ln \frac{\left|x\right|}{\sqrt{x^2+1}}-\frac{3}{2(x^2+1)}+K$
• C. $-\ln \frac{\left|x\right|}{\sqrt{x^2+1}}+\frac{3}{2(x^2+1)}+K$
• D. $-\ln \frac{\left|x\right|}{\sqrt{x^2+1}}+\frac{3}{2(x+1)}+K$

Answer 1

Answer: (A)

$\ln \frac{\left|x\right|}{\sqrt{x^2+1}}+\frac{1}{2(x^2+1)}+K$

Let $I=\int \frac{dx}{x{(x^2+1)}^2}$
$=\int \frac{xdx}{x^2{(x^2+1)}^2}$
Put $x^2+1=t$
$\Rightarrow 2xdx=dt$
$I=\frac{1}{2}\int \frac{1}{t^2(t-1)}dt$
Now, $\frac{1}{t^2(t-1)}=\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t-1}$

$\Rightarrow 1=At(t-1)+B(t-1)+C{t}^2$
Put $t=0\Rightarrow B=-1$
Put $t=1\Rightarrow C=1$
Put $t=-1\Rightarrow 2A-2B+C=1$
$\Rightarrow A=-1$
Hence, the integral becomes
$I=-\frac{1}{2}\int \frac{1}{t}dt-\frac{1}{2}\int \frac{1}{t^2}dt+\frac{1}{2}\int \frac{1}{t-1}dt$
$=-\frac{1}{2}\ln |t|-\frac{1}{2}\frac{-1}{t}+\frac{1}{2}\ln |t-1|+K$
$I=-\frac{1}{2}\ln |x^2+1|+\frac{1}{2}\frac{1}{x^2+1}+\frac{1}{2}\ln |x^2|+K$
$I=\ln \frac{|x|}{\sqrt{x^2+1}}+\frac{1}{2}\frac{1}{x^2+1}+K$