Question

$\int (3.x^2.{\tan }^{-1}x+\frac{x^3}{1+x^2})dx=....+c$.

• A. $x^3{\tan }^{-1}x$
• B. $\frac{x^3}{3}{\tan }^{-1}x$
• C. $x^2{\tan }^{-1}x$
• D. $\frac{x^2}{2}{\tan }^{-1}x$

Answer 1

The correct option is A $x^3{\tan }^{-1}x$

Given $\int (3x^2.{\tan }^{-1}x+\frac{x^3}{1+x^2})dx$= $\int (3x^2.{\tan }^{-1}x.dx)+\int \left(\frac{x^3}{1+x^2}\right)dx$

Integrating by parts

$\int 3x^{2}ta{n}^{-1}x.dx+x^3\int \frac{1}{1+x^2}dx-\int \left(\frac{d}{dx}\right(x^3\left)\right(\int \frac{1}{1+x^2}dx\left)dx\right)$

$\frac{d}{dx}{x}^3=3x^2;\int \frac{1}{1+x^2}=ta{n}^{-1}x$

$\Rightarrow \int 3x^{2}ta{n}^{-1}x.dx+x^3{\tan }^{-1}x-\int 3x^{2}ta{n}^{-1}x.dx$

$\Rightarrow x^3{\tan }^{-1}x+c$