# Antiderivatives of Reciprocal Quadratics $ \begin{align} \int \frac 1 {x^2 + 1} \, dx &= \bigg ( \text {perform trig substitution } x=\tan \theta \bigg) \\ &=\int \frac 1 {\sec^2 \theta} \sec^2 \theta \, d\theta \\ &=\int 1 \, d\theta \\ &=\theta + C \\ &= \arctan(x) +C \end{align} $ $\int \frac 1 {x^2+1} \,dx = \arctan(x) + C$ --- $ \begin{align} \int \frac 1 {x^2 + a^2} \, dx &= \bigg(\text {substitute } x=au, \frac {dx}{du} = a \bigg) \\ &= \int \frac 1 {a^2u^2+a^2} a\,du \\ &= \frac 1 a \int \frac 1 {u^2 + 1} \, du \\ &=\frac 1 a \arctan(u) + C \\ &= \frac 1 a \arctan \left(\frac x a \right) + C \end{align} $ $\int \frac 1 {x^2+a^2} \,dx = \frac 1 a \arctan \left(\frac x a \right) + C$ ^33xemi --- $ \begin{align} \int \frac 1 {(x \pm a)^2 + b^2} \, dx &= \bigg( \text {substitute } u=x\pm a, \frac {du}{dx} =1 \bigg) \\ &= \int \frac 1 {u^2 + b^2} \, du \\ &= \frac 1 b \arctan \left( \frac {x \pm a} b \right) + C \end{align} $ $\int \frac 1 {(x \pm a)^2 + b^2} \, dx = \frac 1 b \arctan \left( \frac {x \pm a} b \right) + C$ ^vkb1ui --- $\int \frac 1 {x^2 - a^2} \, dx = \int \frac A {x-a}\,dx - \int \frac B {x+a} \, dx$ $ \begin{gather} A(x+a) - B(x-a) = 1 \\ Ax-Bx +Aa+Ba = 1 \\ (A-B)x + (A+B)a = 1\\ \end{gather} $ $ \begin{gather} A-B = 0 & (A+B)a = 1\\ A = B & 2A= \frac 1 a \end{gather} $ $A=B=\frac 1 {2a}$ $ \begin{align} \int \frac 1 {x^2 - a^2} \, dx &= \int \frac A {x-a}\,dx - \int \frac B {x+a} \, dx\\ &= \frac 1 {2a}\int \frac 1 {x-a} \, dx - \frac 1 {2a}\int \frac 1 {x+a} \, dx \\ &=\frac 1 {2a} \ln|x-a|-\frac 1 {2a}\ln|x+a| + C \\ &=\frac 1 {2a}\ln \left|\frac {x-a} {x+a}\right| + C \end{align} $ $\int \frac 1 {x^2 - a^2} \, dx =\frac 1 {2a}\ln \left|\frac {x-a} {x+a}\right| + C$ ^6w7yvr --- $ \begin{align} \int \frac 1 {(x \pm a)^2 - b^2} \, dx &= \bigg( \text {substitute } u=x\pm a, \frac {du}{dx} =1 \bigg) \\ &= \int \frac 1 {u^2 - b^2} \, du \\ &= \frac 1 {2b}\ln \left|\frac {u-b} {u+b}\right| + C \end{align} $ $\int \frac 1 {(x \pm a)^2 - b^2} \, dx = \frac 1 {2b}\ln \left|\frac {x \pm a -b} {x \pm a+b}\right| + C$ ^w2nryz --- $ \begin{align} \int \frac 1 {(x\pm a)^2} \, dx &= \bigg( \text {substitute } u=x \pm a, \frac {du}{dx} =1 \bigg) \\ &= \int u^{-2} \, du \\ &= -u^{-1} + C \\ &=- \frac 1 {x \pm a} + C \end{align} $ $\int \frac 1 {(x\pm a)^2} \, dx =- \frac 1 {x \pm a} + C$ ^u28w30 ## Sources - [Integrating the reciprocal of a quadratic](https://www.youtube.com/watch?v=f-dzLF98cys)