∫ x ^ 2 √( 1 x ) #160; dx Let, u=1+x =∫( u 1 ) #160; ^ 2 √( u ) #160; du =∫ u ^ 3 2 #160; 2√( u ) + 1 √( u ) #160; #16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration by Substitution

Solve: ∫x2√1...

Question

Solve: $\int \frac{x^{2}}{\sqrt{1 - x}} . d x$

Open in App

Solution

$\int \frac{x^{2}}{\sqrt{1 - x}} d x L e t, u = 1 + x = \int \frac{{(u - 1)}^{2}}{\sqrt{u}} d u = \int u^{\frac{3}{2}} - 2 \sqrt{u} + \frac{1}{\sqrt{u}} d u = \int u^{\frac{3}{2}} d u - \int 2 \sqrt{u} d u + \int \frac{1}{\sqrt{u}} d u = \frac{2}{5} {(1 + x)}^{\frac{5}{2}} - \frac{4}{3} {(1 + x)}^{\frac{3}{2}} + 2 \sqrt{1 + x} + C$

Suggest Corrections

Similar questions

\int_{0}^{\infty} ℓ n (x + \frac{1}{x}) . \frac{d x}{1 + x^{2}}

\int \frac{\sqrt{1 - \sqrt{x}}}{\sqrt{1 + \sqrt{x}}} . d x

\int \frac{sin x}{1 - 4 {cos}^{2} x} . d x .

\int \sqrt{tan x} + \sqrt{cot x} . d x

\int_{0}^{1} \frac{1 - x}{1 + x} . \frac{d x}{\sqrt{x + x^{2} + x^{3}}}

Join BYJU'S Learning Program