int frac{sqrt{x-2}}{2x+4} dx= | vedclass

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

Question

(B) Let $I = \int \frac{\sqrt{x-2}}{2x+4} dx$. Substitute $t = \sqrt{x-2}$,then $t^2 = x-2$,so $x = t^2+2$ and $dx = 2t dt$. Substituting these into t

Vedclass · Accepted Answer

$\sqrt{x-2}-2 \operatorname{Tan}^{-1}\left(\frac{\sqrt{x-2}}{2}\right)+c$

Vedclass · Answer

(B) Let $I = \int \frac{\sqrt{x-2}}{2x+4} dx$. Substitute $t = \sqrt{x-2}$,then $t^2 = x-2$,so $x = t^2+2$ and $dx = 2t dt$. Substituting these into the integral: $I = \int \frac{t}{2(t^2+2)+4} (2t dt) = \int \frac{2t^2}{2t^2+8} dt = \int \frac{t^2}{t^2+4} dt$. Now,rewrite the integrand: $I = \int \frac{t^2+4-4}{t^2+4} dt = \int \left(1 - \frac{4}{t^2+4}\right) dt$. Integrating term by term: $I = \int 1 dt - 4 \int \frac{1}{t^2+2^2} dt = t - 4 \left(\frac{1}{2} \operatorname{Tan}^{-1}\left(\frac{t}{2}\right)\right) + c$. $I = t - 2 \operatorname{Tan}^{-1}\left(\frac{t}{2}\right) + c$. Substituting $t = \sqrt{x-2}$ back: $I = \sqrt{x-2} - 2 \operatorname{Tan}^{-1}\left(\frac{\sqrt{x-2}}{2}\right) + c$.

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

Similar Questions

Find the integral of the function $\frac{\cos x-\sin x}{1+\sin 2 x}$.

The integral $\int \frac{dx}{(x+4)^{\frac{8}{7}}(x-3)^{\frac{6}{7}}}$ is equal to (where $C$ is a constant of integration).

The integral $\int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$,where $x > 0$,is equal to ....... (where $c$ is a constant of integration).

If $f(x) = \sqrt{\tan x}$ and $g(x) = \sin x \cdot \cos x$,then $\int \frac{f(x)}{g(x)} dx$ is equal to (where $C$ is a constant of integration).

If ${I_1} = \int_e^{{e^2}} \frac{dx}{\log x}$ and ${I_2} = \int_1^2 \frac{e^x}{x} dx$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module