Question

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

• A. $\frac{1}{3}\log |x+1|-\frac{1}{6}\log |x^2-x+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+c$
• B. $\frac{1}{3}\log |x+1|+\frac{1}{6}\log |x^2-x+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+c$
• C. $\frac{1}{3}\log |x+1|-\frac{1}{6}\log |x^2-x+1|-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+c$
• D. $-\frac{1}{3}\log |x+1|+\frac{1}{6}\log |x^2-x+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+c$

Answer 1

Answer: (A)

$\frac{1}{3}\log |x+1|-\frac{1}{6}\log |x^2-x+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+c$

$I=\int \frac{1}{1+x^3}dx$
Let $\frac{1}{1+x^3}=\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}$,
On comparing the terms we get,
$A=\frac{1}{3}$ , $C=\frac{2}{3}$ and $B=-\frac{1}{3}$
Hence, $I=\int \frac{1}{3(x+1)}dx+\int \frac{\frac{-x}{3}+\frac{2}{3}}{x^2-x+1}$
$I=\frac{1}{3}log|x+1|-\frac{1}{6}\int \frac{2x-1}{x^2-x+1}dx+\frac{1}{2}\int \frac{dx}{x^2-x+1}$
$I=\frac{1}{3}log|x+1|-\frac{1}{6}log|x^2-x+1|+\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$
Hence, option A is correct