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

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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}} = $

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