Mathematics

# Evaluate the following integral : $\displaystyle \int \sqrt{4x^{2}+9} \ dx$

$\dfrac{x}{2}.\sqrt{4x^{2}+9}+\dfrac{9}{4} \ln \left | 2x+\sqrt{4x^{2}+9} \right |+C$

##### SOLUTION
Let $I=\displaystyle\int \sqrt{4 x^2+9}d x$

$\implies I=\displaystyle\int \sqrt{4\left(x^2+\dfrac{9}{4}\right)}d x$

$\implies I=2\displaystyle\int \sqrt{x^2+\left(\dfrac{3}{2}\right)^2} d x$

As we know that

$\displaystyle\int \sqrt{a^2+x^2} d x=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|+C$

Here $a=\dfrac{3}{2}$

$\implies I=2\left(\dfrac{x}{2}\sqrt{x^2+\dfrac{9}{4}}+\dfrac{9/4}{2}\ln \left|x+\sqrt{x^2+\dfrac{9}{4}}\right|\right)+C$

$\implies I=\dfrac{x}{2}\sqrt{4\bigg(x^2+\dfrac{9}{4}\bigg)}+\dfrac{9}{4}\ln \left|x+\sqrt{\dfrac{4 x^2+9}{4}}\right|+C$

$\implies I=\dfrac{x}{2}\sqrt{4 x^2+9}+\dfrac{9}{4}\ln \left|\dfrac{2 x}{2}+\dfrac{\sqrt{ 4 x^2+9}}{2}\right|+C$

$\implies I=\dfrac{x}{2}\sqrt{4 x^2+9}+\dfrac{9}{4}\ln \left|2 x+\sqrt{4 x^2+9}\right|+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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Evaluate: $\displaystyle \int e^{2x} \sin 3x \ dx$

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Evaluate: $\displaystyle \int { \cfrac { \sec ^{ 2 }{ x } }{ \tan ^{ 2 }{ x } +4 } } dx\quad$

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$\displaystyle \int { \frac { x+\sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 6 ]{ x } }{ x\left( 1+\sqrt [ 3 ]{ x } \right) } dx }$ is equal to
• A. $\displaystyle \frac { 3 }{ 2 } { x }^{ 2/3 }-6\tan ^{ -1 }{ { x }^{ 1/6 } } +c$
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$\displaystyle\int \frac{x+2}{\sqrt{\left ( 4x-x^{2} \right )}}dx.$
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Consider the integrals $I_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{tan x}}$ and $I_2=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{sin x}dx}{\sqrt{sin }x+\sqrt{cos}x}$