Question

Integrate:

$\int \frac{x^4}{x^2+1}$ dx =

• A. $\frac{x^3}{3}-x+{\tan }^{-1}x+c$
• B. $\frac{x^5}{5}+{\tan }^{-1}x+c$
• C. $4x^3+{\tan }^{-1}x+c$
• D. $\frac{x^4}{4}-x+{\tan }^{-1}x+c$

Answer 1

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

$\int \frac{x^4}{x^2+1}dx=\int \frac{x^4+1-1}{x^2+1}dx=\int (\frac{x^4-1}{x^2+1}+\frac{1}{x^2+1})dx$

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

$=\frac{x^3}{3}-x+ta{n}^{-1}x+c$