Question

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

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

Answer 1

Answer: (B)

Let $I=\int \frac{x^3-1}{x^3+x}dx=\int \left(1-\frac{x+1}{x^3+x}\right)dx$
$=\int 1dx-\int \frac{x+1}{x(x^2+1)}dx=x-\int \frac{x+1}{x(x^2+1)}$ ....(i)
Now $\frac{x+1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$
(By using partial fractions)
$\Rightarrow x+1=A(x^2+1)+(Bx+C)x$
$\Rightarrow x+1=(A+B)x^2+Cx+A$
Comparing coefficients of $x^2,x$ and constant,
we get
$A+B=0,C=1,A=1$
$\Rightarrow B=-1$
$\therefore$ From (i), we get
$I=x-\int \frac{1}{x}dx-\int \frac{1-x}{x^2+1}dx$
$=x-\log x-\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{2x}{x^2+1}dx$
$=x-\log x-{\tan }^{-1}x+\frac{1}{2}\log (x^2+1)+c$