Question

$\int \frac{x^3-1}{x^3+x}dx$ is equal to

• A. $x-\log x+\log (x^2+1)-{\tan }^{-1}x+c$
• B. $x-\log x+\frac{1}{2}\log (x^2+1)-{\tan }^{-1}x+c$
• C. $x+\log x+\frac{1}{2}\log (x^2+1)+{\tan }^{-1}x+c$
• D. $x+\log x-\frac{1}{2}\log (x^2+1)-{\tan }^{-1}x+c$

Answer 1

Answer: (B)

$x-\log x+\frac{1}{2}\log (x^2+1)-{\tan }^{-1}x+c$

$I=\int \frac{x^3-1}{x^3+x}dx$

$I=\int [1-\frac{(x+1)}{(x^3+x)}]dx$

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

Now, $\frac{x+1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$ .....(1)

$\Rightarrow x+1=A(x^2+1)+(Bx+C)x$

$A+B=0$

$A=1,B=-1,C=1$

Put these values in (1), we get
$\frac{x+1}{x(x^2+1)}=\frac{1}{x}+\frac{-x+1}{x^2+1}$

$\Rightarrow I=x-\int \frac{1}{x}+\frac{-x+1}{x^2+1}dx$

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