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# Evaluate the following integrals:

Question:

Evaluate the following integrals: $\int \frac{x^{2}}{\sqrt{1-x}} d x$

Solution:

$\operatorname{Let} I=\int \frac{x^{2}}{\sqrt{1-x}} d x$

Substituting $1-x=t \Rightarrow d x=-d t$,

$\Rightarrow \mathrm{I}=-\int \frac{(1-\mathrm{t})^{2}}{\sqrt{\mathrm{t}}} \mathrm{dt}$

$\Rightarrow \mathrm{I}=-\int \frac{\mathrm{t}^{2}-2 \mathrm{t}+1}{\sqrt{\mathrm{t}}} \mathrm{dt}$

$\Rightarrow \mathrm{I}=-\int\left(\mathrm{t}^{\frac{3}{2}}-2 \mathrm{t}^{\frac{1}{2}}+\mathrm{t}^{-\frac{1}{2}}\right) \mathrm{dt}$

$\Rightarrow I=-\left[\frac{2}{5} t^{\frac{5}{2}}+2 t^{\frac{1}{2}}-\frac{4}{3} t^{\frac{3}{2}}\right]+c$

$\Rightarrow I=\frac{-\left(6 t^{\frac{5}{2}}+30 t^{\frac{1}{2}}-20 t^{\frac{3}{2}}\right)}{15}+c$

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

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

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

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

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