Evaluate the definite integral $\int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(D) Let $I = \int_{0}^{1} \frac{dx}{\sqrt{1+x}-\sqrt{x}}$.
Rationalizing the denominator:
$I = \int_{0}^{1} \frac{1}{(\sqrt{1+x}-\sqrt{x})} \times \frac{(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})} dx$
$I = \int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{(1+x)-x} dx$
$I = \int_{0}^{1} (\sqrt{1+x} + \sqrt{x}) dx$
$I = \left[ \frac{2}{3}(1+x)^{3/2} \right]_{0}^{1} + \left[ \frac{2}{3}x^{3/2} \right]_{0}^{1}$
$I = \frac{2}{3} \left[ (1+1)^{3/2} - (1+0)^{3/2} \right] + \frac{2}{3} [1^{3/2} - 0^{3/2}]$
$I = \frac{2}{3} [2^{3/2} - 1] + \frac{2}{3} [1]$
$I = \frac{2}{3} (2\sqrt{2} - 1) + \frac{2}{3}$
$I = \frac{4\sqrt{2}}{3} - \frac{2}{3} + \frac{2}{3} = \frac{4\sqrt{2}}{3}$.

Explore More

Similar Questions

$\int_0^1 x \left|x - \frac{1}{2}\right| dx = $

$\int_0^1 {(1 - x)^9} dx = $

Using the Trapezoidal rule, find the approximate value of $\int_1^4 y \, dx$ based on the following data:
$x$$1$$2$$3$$4$
$y$$0.7111$$0.7222$$0.7333$$0.7444$
(in $.1833$)

$\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=$

$\int_0^4 \frac{x+2}{\sqrt{4x-x^2}} dx =$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo