int frac{dx}{sqrt{x + a} + sqrt{x + b}} = | vedclass

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

Question

(B) To solve the integral $I = \int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}}$,we rationalize the denominator by multiplying the numerator and denominato

Vedclass · Accepted Answer

$\frac{2}{3(a - b)}[(x + a)^{3/2} - (x + b)^{3/2}] + C$

Vedclass · Answer

(B) To solve the integral $I = \int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}}$,we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{x + a} - \sqrt{x + b})$.$I = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{(x + a) - (x + b)} dx$$I = \int \frac{\sqrt{x + a} - \sqrt{x + b}}{a - b} dx$Since $(a - b)$ is a constant,we can take it out of the integral:$I = \frac{1}{a - b} \int ((x + a)^{1/2} - (x + b)^{1/2}) dx$Using the power rule for integration $\int u^n du = \frac{u^{n+1}}{n+1} + C$:$I = \frac{1}{a - b} [\frac{(x + a)^{3/2}}{3/2} - \frac{(x + b)^{3/2}}{3/2}] + C$$I = \frac{2}{3(a - b)} [(x + a)^{3/2} - (x + b)^{3/2}] + C$Thus,the correct option is $B$.

$\int \frac{dx}{\sqrt{x + a} + \sqrt{x + b}} = $

Similar Questions

$\int \frac{d x}{x^2+2 x+5} = $ . . . . . . $+ C$.

If $f(x) = \frac{3-8x}{3x-1}$ and $\int f(y) dy = Ay + B \log |3y-1| + C$, then $\frac{A-3B}{2} =$

The solution of the equation $\frac{d^2y}{dx^2} = e^{-2x}$ is

$\int \frac{x^3-7 x+6}{x^2+3 x} \,d x=$

Find the integral of the function $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module