Question

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

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

$I=\int \frac{x^3-1}{x^3+x}dxI=\int \frac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2+1\right)}dxI=\int \frac{x-1}{x}+\int \frac{x-1}{x^2+1}dx$

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

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

$I=x-{\log }_{e}x+\frac{1}{2}{\log }_e\left(x^2+1\right)-{\tan }^{-1}x+C$

Hence, this is the answer.