Test Index

Indefinite Integration

Section: Mathematics

Question : 7 of 89

Marks: +1, -0

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

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

Solution:

\;\; \text{ Let } \; I = \int \frac{\sqrt{x-2}}{2x+4} dx

\;\; \text{ and } \; u = \sqrt{x-2}

\; \Rightarrow u^2 = x-2 \; \text{ and } \; x = u^2+2

Differentiate

x

w.r.t

u

, we get

\; \frac{dx}{du} \; = 2u \Rightarrow dx = 2u \, du

I\; = \int \frac{\sqrt{x-2}}{2x+4} dx = \int \frac{2u \cdot u}{2(u^2+2)+4} du

\; = \int \frac{2u^2}{2u^2+8} du = \int \frac{u^2}{u^2+4} du

\; = \int \frac{u^2+4-4}{u^2+4} du = \int \left(1 - \frac{4}{u^2+4}\right) du

\; = u - 4 \cdot \; \frac{1}{2} \cdot \tan^{-1}\left( \frac{u}{2} \right) + C

where,

C=

constant

\; = u - 2 \tan^{-1}\left( \frac{u}{2} \right) + C

\; = \sqrt{x-2} - 2 \tan^{-1}\left( \frac{\sqrt{x-2}}{2} \right) + C

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