Question

if $\int \frac{x^4+1}{x(x^2+1{)}^2}dx=Aln|x|+\frac{B}{1+x^2}+c$, where c is the constant of integration then :

• A. $A=1,B=-1$
• B. $A=-1,B=1$
• C. $A=1,B=1$
• D. $A=-1,B=-1$

Answer 1

The correct option is C $A=1,B=1$

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

$=ln|x|-\int \left(\frac{2x}{(x^2+1{)}^2}\right)dx=ln|x|-I_1$

$I_1=\int \left(\frac{2x}{(x^2+1{)}^2}\right)dx$

Put $x^2+1=t⟹2xdx=dt$

$I_1=\int t^{-2}dt=\frac{1}{t}=\frac{1}{x^2+1}$

$\therefore I=ln|x|+\frac{1}{x^2+1}+c$

Comparing with $Aln|x|+\frac{B}{x^2+1}+c$

We get,

$A=1,B=1$