Question

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

Answer 1

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

$u=\sqrt{x}$

$dx=2\sqrt{x}dx$

$x=u^2$

$x^{3/2}=u^3$

$x^2=u^4$

$\Rightarrow 2\int \frac{(u-1)u^2{(u+1)}^2}{u^2+u+1}du$

Perform polynomial long division

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

$=2\int \frac{u-1}{u^2+u+1}du+2\int u^{3}du-\int udu+2\int du$

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

$\Rightarrow \int \frac{2u+1}{u^2+u+1}du-3\int \frac{du}{u^2+u+1}+\frac{{2u}^4}{4}-\frac{{4u}^2}{2}+2u+c$

Putting $u^2+u+1=t$

$(2u+1)du=dt$

$\Rightarrow 1\int \frac{dt}{t}-3\int \frac{du}{[u^2+u+\frac{1}{4}]+3/4}+\frac{u^4}{2}-2u^2+2u+c$

$\Rightarrow Int-3\times \frac{2}{\sqrt{3}}{tan}^{-1}\left[\frac{(u+\frac{1}{2})}{\sqrt{3}/2}\right]+\frac{u^4}{2}-2u^2+2u+c$

$\Rightarrow In(u^2+u+1)-2\sqrt{3}{tan}^{-1}\left(\frac{2u+1}{\sqrt{3}}\right)+\frac{u^4}{2}-2u^2+2u+c$

$\Rightarrow -2\sqrt{3}{tan}^{-1}\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)+In(x+\sqrt{x}+1)+\frac{(x-4)x}{2}+2\sqrt{x}+1$