Mathematics

# Evaluate:$\displaystyle\int\dfrac{x}{\sqrt{3x^2+4}}dx$.

##### SOLUTION
Now,
$\displaystyle\int\dfrac{x}{\sqrt{3x^2+4}}dx$
$=\dfrac{1}{6}\displaystyle\int\dfrac{6x}{\sqrt{3x^2+4}}dx$
$=\dfrac{1}{6}\displaystyle\int\dfrac{d(3x^2+4)}{\sqrt{3x^2+4}}dx$
$=\dfrac{1}{6}.2.\sqrt{3x^2+4}+c$ [ Where $c$ is integrating constant]
$=\dfrac{1}{3}\sqrt{3x^2+4}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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