int frac{x}{sqrt{x+4}} , dx = . . . . . . | vedclass

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

Question

(B) To evaluate the integral $I = \int \frac{x}{\sqrt{x+4}} \, dx$,let $u = x+4$. Then $du = dx$ and $x = u-4$. Substituting these into the integral,w

Vedclass · Accepted Answer

$\frac{2}{3} \sqrt{x+4}(x-8)$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{x}{\sqrt{x+4}} \, dx$,let $u = x+4$. Then $du = dx$ and $x = u-4$. Substituting these into the integral,we get: $I = \int \frac{u-4}{\sqrt{u}} \, du$ $I = \int (\frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}}) \, du$ $I = \int (u^{1/2} - 4u^{-1/2}) \, du$ Integrating term by term: $I = \frac{u^{3/2}}{3/2} - 4 \frac{u^{1/2}}{1/2} + C$ $I = \frac{2}{3} u^{3/2} - 8 u^{1/2} + C$ $I = \frac{2}{3} u^{1/2} (u - 12) + C$ Substituting $u = x+4$ back: $I = \frac{2}{3} \sqrt{x+4} (x+4 - 12) + C$ $I = \frac{2}{3} \sqrt{x+4} (x-8) + C$ Thus,the correct option is $B$.

$\int \frac{x}{\sqrt{x+4}} \, dx = $ . . . . . . $+ C, x > -4$.

Similar Questions

$\int \frac{x^{e-1}+e^{x-1}}{x^e+e^x} d x=$

$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx = $ . . . . . . $+ C$.

$\int \frac{\sec^{8} x}{\text{cosec} x} dx =$

$\int \frac{dx}{1+\sqrt{x}} = $

If $l^r(x)$ denotes the $r$-th iterated logarithm of $x$,i.e.,$l^1(x) = \log(x)$,$l^2(x) = \log(\log(x))$,...,$l^r(x) = \log(\log(...\log(x)...))$,then $\int \frac{1}{x \cdot l^1(x) \cdot l^2(x) \cdot ... \cdot l^r(x)} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module