Mathematics

Solve $$\displaystyle\int \dfrac{\sqrt{1+x^2}}{x^2}dx$$


SOLUTION

Given: $$ \displaystyle\int \dfrac{\sqrt{1+x^2}}{x^2}dx $$

Apply the integration by parts,

$$ u = \sqrt{1 + x^{2}}, v = \dfrac{1}{x^{2}} $$

$$ \displaystyle\int \dfrac{\sqrt{1+x^2}}{x^2}dx = -\dfrac{\sqrt{1+x^2}}{x}-\int \:-\dfrac{1}{\sqrt{1+x^2}}dx $$

$$ = -\dfrac{\sqrt{1+x^2}}{x}-\left(-\ln \left|\sqrt{x^2+1}+x\right|\right) $$

$$ = -\dfrac{\sqrt{1+x^2}}{x}+\ln \left|\sqrt{x^2+1}+x\right|+C $$

Hence, the required result is found.

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Subjective Medium Published on 17th 09, 2020
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