Question

Solve : $\int \frac{x^2}{(4+x{)}^{3/2}}dx$

Answer 1

$I=\int \frac{x^2}{(4+x{)}^{\frac{3}{2}}}dx$

$(4+x)=t\Rightarrow x=(t-4)$

$dx=dt$

$=I=\int \frac{(t-4{)}^{2}dt}{t^{\frac{3}{2}}}$

$=\int \frac{t^2-8^{\frac{1}{2}}+16}{t^{\frac{3}{2}}}dt$

$=\int [t^{\frac{1}{2}}-\frac{8}{t^{\frac{1}{2}}}+t^{\frac{-3}{2}}]dt$

$\Rightarrow \left(\frac{t^{\frac{1}{2}}+1}{\frac{1}{2}+1}\right)-8\left(\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\right)+\frac{t^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+c$

$\Rightarrow \frac{2t^{\frac{3}{2}}}{3}-1b{t}^{\frac{1}{2}}-t\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+c$

$\Rightarrow \frac{2t^{\frac{3}{2}}}{3}-16t^{\frac{1}{2}}-2t^{\frac{1}{2}}+c$

$\Rightarrow 2[\frac{t^{\frac{3}{2}}}{3}-8t^{\frac{1}{2}}-t^{\frac{-1}{2}}]+c$

$\Rightarrow 2[\frac{(4+x{)}^{\frac{3}{2}}}{3}-8(4+x{)}^{\frac{1}{2}}-\frac{1}{(4+x{)}^{\frac{1}{2}}}]+c$

$\Rightarrow 2\left[\frac{(4+x{)}^2-(8.3\left)\right(4+x)-1}{3(4+x{)}^{\frac{1}{2}}}\right]+c$

$\Rightarrow \frac{2}{3}\left[\frac{x^2+8x+16-96-24x-1}{(x+4{)}^{\frac{1}{2}}}\right]+c$

$I\Rightarrow \frac{2}{3}\left[\frac{x^2-16x-81}{(x+4{)}^{\frac{1}{2}}}\right]$