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
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