Question

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

• A. $\frac{x^4}{4}+\frac{x^3}{3}-{\tan }^{-1}x+c$
• B. $\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2}\log (x^2+1)+c$
• C. $\frac{x^4}{4}+\frac{x^3}{3}+{\tan }^{-1}x+c$
• D. $\frac{x^4}{4}+\frac{x^3}{5}-{\tan }^{-1}x+c$

Answer 1

Answer: (B)

$\frac{x^4}{4}-\frac{x^2}{2}+\frac{1}{2}\log (x^2+1)+c$

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

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

Put $x^2+1=t$

$2xdx=dt$

$=\frac{1}{2}\int \frac{{\left(t-1\right)}^2}{t}dt$

$=\frac{1}{2}\int \left(t+\frac{1}{t}-2\right)dt$

$=\frac{1}{2}\int \left(\frac{t^2}{2}+\log t-2t\right)tc$

$=\frac{{\left(x^2+1\right)}^2}{4}+\frac{1}{2}\log \left(x^2+1\right)-1\left(x^2+1\right)tc$

$=\frac{x^4+1+2x^2}{4}-x^2-1+\frac{1}{2}\log \left(x^2+1\right)tc$

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