Prove that: ∫ dfracmathrmdx√(x²+a²) = log|x+√(x²+a²)|+c

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\displaystyle\int \dfrac{\mathrm{d}x}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+c

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2019Subjective· 4mImportance★★★★★

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Substitute $x=a\tan\theta$ to reduce the integral to $\int\sec\theta\,d\theta$ .

Let $I = \displaystyle\int\dfrac{dx}{\sqrt{x^2+a^2}}$ .

Put $x = a\tan\theta \implies dx = a\sec^2\theta\,d\theta$ , and $\sqrt{x^2+a^2} = \sqrt{a^2\tan^2\theta+a^2} = a\sec\theta$ (taking $\sec\theta>0$ ).

$I = \int \dfrac{a\sec^2\theta\,d\theta}{a\sec\theta} = \int \sec\theta\,d\theta = \log|\sec\theta+\tan\theta| + c$

Now, since $\tan\theta = \dfrac xa$ and $\sec\theta = \dfrac{\sqrt{x^2+a^2}}{a}$ :

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Q.Prove that: $\displaystyle\int \dfrac{\mathrm{d}x}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+c$

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Q.Prove that: ∫dxx2+a2=log⁡∣x+x2+a2∣+c\displaystyle\int \dfrac{\mathrm{d}x}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+c∫x2+a2​dx​=log​x+x2+a2​​+c

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Q.Prove that: $\displaystyle\int \dfrac{\mathrm{d}x}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+c$

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