Question

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

• A. $\frac{1}{15}\sqrt{1+x^2}\left(3x^4+4x^2+8\right)+C$
• B. $\frac{1}{15}\sqrt{1+x^2}\left(3x^4-4x^2+8\right)+C$
• C. $\sqrt{1+x^2}\left(3x^4+4x^2+8\right)+C$
• D. None of these

Answer 1

The correct option is B $\frac{1}{15}\sqrt{1+x^2}\left(3x^4-4x^2+8\right)+C$

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

Let,

$1+x^2=x$

$x^4={\left(u-1\right)}^2$

$2xdx=du$

$C=\frac{1}{2}\int \frac{{\left(u-1\right)}^2}{\sqrt{u}}du$

$=\frac{1}{2}\int \frac{u^2+1-2u}{\sqrt{u}}du$

$=\frac{1}{2}\int \left(u^{\frac{3}{2}}+u^{\frac{-1}{2}}-2u^{\frac{1}{2}}\right)du$

$=\frac{1}{2}\left(\frac{u^{\frac{5}{2}}}{\frac{5}{2}}+2u^{\frac{1}{2}}-\frac{2u^{\frac{3}{2}}}{\frac{3}{2}}\right)+c$

$=\frac{1}{5}{u}^{\frac{5}{2}}+u^{\frac{1}{2}}-\frac{2u^{\frac{3}{2}}}{3}+c$

$=\frac{u^{\frac{1}{2}}}{15}\left(3u^2+15-10u\right)+c$

$=\frac{\sqrt{1+x^2}}{15}\left[3\left(1+x^4+2x^2\right)+15-10\left(1+x^2\right)\right]+c$

$=\frac{\sqrt{1+x^2}}{15}\left(3+3x^4+6x^2+15-10-10x^2\right)+c$

$=\frac{\sqrt{1+x^2}}{15}\left(3x^4-4x^2+8\right)+c$

(B) option.