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Integral with x2±a2x^2 \pm a^2

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∫1x2+a2 dx=1aarctan⁡(xa)+C\int\frac{1}{x^2+a^2}\ \mathrm{d}x=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C
Derivation
I=∫1x2+a2 dx=∫1a2(x2a2+1) d(xa)1a(Direct Substitution)=1a∫1(xa)2+1 d(xa)=1aarctan⁡(xa)+C\begin{align*} I&=\int{\frac{1}{x^2+a^2}}\ \mathrm{d}x\\ &=\int{\frac{1}{a^2\left(\frac{x^2}{a^2}+1\right)}}\ \frac{\mathrm{d}\left(\frac{x}{a}\right)}{\frac{1}{a}} &&\text{(Direct Substitution)}\\ &=\frac{1}{a}\int\frac{1}{\left(\frac{x}{a}\right)^2+1}\ \mathrm{d}\left(\frac{x}{a}\right)\\ &=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C \tag*{$\blacksquare$} \end{align*}

Direct Substitution

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∫1(x2+a2)n dx=x2(n−1)a2(x2+a2)n−1+2n−32(n−1)a2∫1(x2+a2)n−1 dx\int \frac{1}{\left(x^2+a^2\right)^n}\ \mathrm{d}x=\frac{x}{2\left(n-1\right)a^2\left(x^2+a^2\right)^{n-1}}+\frac{2n-3}{2\left(n-1\right)a^2}\int \frac{1}{\left(x^2+a^2\right)^{n-1}}\ \mathrm{d}x
Derivation

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∫1x2−a2 dx=12aln⁡∣x−ax+a∣+C\int \frac{1}{x^2-a^2}\ \mathrm{d}x=\frac{1}{2a}\ln{\left| \frac{x-a}{x+a}\right|}+C
Derivation
I=∫1x2−a2 dx=∫x+a−(x−a)2a(x2−a2) dx=12a[∫(x+a)−(x−a)(x+a)(x−a) dx]=12a(∫1x−a dx−∫1x+a dx)=12a(ln⁡∣x−a∣−ln⁡∣x+a∣+C)=12aln⁡∣x−ax+a∣+C\begin{align*} I&=\int{\frac{1}{x^2-a^2}}\ \mathrm{d}x\\ &=\int{\frac{x+a-(x-a)}{2a(x^2-a^2)}}\ \mathrm{d}x\\ &=\frac{1}{2a}\left[\int{\frac{(x+a)-(x-a)}{(x+a)(x-a)}}\ \mathrm{d}x \right]\\ &=\frac{1}{2a}\left(\int{\frac{1}{x-a}}\ \mathrm{d}x-\int{\frac{1}{x+a}}\ \mathrm{d}x\right)\\ &=\frac{1}{2a}\left(\ln{\left|x-a\right|}-\ln{\left|x+a\right|}+C\right)\\ &=\frac{1}{2a}\ln{\left| \frac{x-a}{x+a}\right|}+C \tag*{$\blacksquare$} \end{align*}