Prove that $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a} an^{-1}\frac{x}{a} + c$.

Question

Prove that $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}	an^{-1}\frac{x}{a} + c$.

Accepted Answer

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> [!TLDR]
> Substitute $x=a	an	heta$ to reduce the integrand to a constant.
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> Let $x=a	an	heta$, so that $dx=a\sec^2	heta\,d	heta$ and
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> $$x^2+a^2=a^2	an^2	heta+a^2=a^2(	an^2	heta+1)=a^2\sec^2	heta$$
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> Substituting,
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> $$\int\frac{1}{x^2+a^2}\,dx=\int\frac{a\sec^2	heta}{a^2\sec^2	heta}\,d	heta=\int\frac1a\,d	heta=\frac{	heta}{a}+c$$
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Q.Prove that $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + c$ .

Concept understanding — Standard Integral Forms

Q.Prove that ∫1x2+a2 dx=1atan⁡−1xa+c\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + c∫x2+a21​dx=a1​tan−1ax​+c.

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Q.Prove that $\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + c$ .

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