Prove that $$\int \sqrt{x^2+a^2}\,dx = \dfrac{1}{2}\left[x\sqrt{x^2+a^2} + a^2 \log|x + \sqrt{x^2+a^2}|\right] + c.$$

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Accepted Answer

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> [!TLDR]
> integrate by parts with u=√(x²+a²), dv=dx, then solve the resulting equation for the same integral I
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> Let $I=\displaystyle\int\sqrt{x^2+a^2}\,dx$. Integrate by parts with $u=\sqrt{x^2+a^2}$, $dv=dx$ (so $v=x$, $du=\dfrac{x}{\sqrt{x^2+a^2}}dx$):
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> $$I=x\sqrt{x^2+a^2}-\int\dfrac{x^2}{\sqrt{x^2+a^2}}\,dx$$
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> Write $x^2=(x^2+a^2)-a^2$:
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> $$\int\dfrac{x^2}{\sqrt{x^2+a^2}}dx=\int\sqrt{x^2+a^2}\,dx-a^2\int\dfrac{dx}{\sqrt{x^2+a^2}}=I-a^2\ln\left|x+\sqrt{x^2+a^2}\right|$$
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> (using the standard result $\int\frac{dx}{\sqrt{x^2+a^2}}=\ln|x+\sqrt{x^2+a^2}|$). So
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Q.Prove that $\int \sqrt{x^2+a^2}\,dx = \dfrac{1}{2}\left[x\sqrt{x^2+a^2} + a^2 \log|x + \sqrt{x^2+a^2}|\right] + c.$

Concept understanding — Standard Integral Forms

Q.Prove that ∫x2+a2 dx=12[xx2+a2+a2log⁡∣x+x2+a2∣]+c.\int \sqrt{x^2+a^2}\,dx = \dfrac{1}{2}\left[x\sqrt{x^2+a^2} + a^2 \log|x + \sqrt{x^2+a^2}|\right] + c.∫x2+a2​dx=21​[xx2+a2​+a2log∣x+x2+a2​∣]+c.

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Q.Prove that $\int \sqrt{x^2+a^2}\,dx = \dfrac{1}{2}\left[x\sqrt{x^2+a^2} + a^2 \log|x + \sqrt{x^2+a^2}|\right] + c.$

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