Evaluate the definite integral: int_0^1 frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+x} + \sqr

Vedclass · Accepted Answer

$\frac{4\sqrt{2}}{3}$

Vedclass · Answer

(B) Let $I = \int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$.Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+x} + \sqrt{x})$:$I = \int_0^1 \frac{\sqrt{1+x} + \sqrt{x}}{(1+x) - x} dx = \int_0^1 (\sqrt{1+x} + \sqrt{x}) dx$.Integrating term by term:$I = \left[ \frac{2}{3}(1+x)^{3/2} + \frac{2}{3}x^{3/2} \right]_0^1$.Evaluating at the limits:$I = \left( \frac{2}{3}(2)^{3/2} + \frac{2}{3}(1)^{3/2} \right) - \left( \frac{2}{3}(1)^{3/2} + 0 \right)$.$I = \frac{2}{3}(2\sqrt{2}) + \frac{2}{3} - \frac{2}{3} = \frac{4\sqrt{2}}{3}$.

Evaluate the definite integral: $\int_0^1 \frac{dx}{\sqrt{1+x} - \sqrt{x}}$

Similar Questions

$\int_0^a x^2 (a^2 - x^2)^{3/2} dx = $

The value of the definite integral $\int\limits_0^{\frac{1}{{\sqrt 2 }}} {\frac{{{x^2}dx}}{{\sqrt {1 - {x^2}} \,(1 + \sqrt {1 - {x^2}} )}}} $ is

$\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$ is equal to :

$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} \,d x =$

$\int_{a-c}^{b-c} f(x+c) \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module