Question

Solve $\int \frac{x^5}{x^2+1}dx$

• A. $\frac{x^4}{4}+\frac{x^2}{2}+ta{n}^{-1}x+c$
• B. $\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2}\log \left(x^2+1\right)+c$
• C. $\frac{x^4}{4}+\frac{x^3}{3}+ta{n}^{-1}x+c$
• D. $\frac{x^4}{4}-\frac{x^3}{3}-ta{n}^{-1}+c$

Answer 1

The correct option is B $\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2}\log \left(x^2+1\right)+c$

Consider the given integral.

$I=\int \frac{x^5}{x^2+1}dx$

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

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

$I=\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2}{\log }_e(x^2+1)+C$

Hence, this is the answer.