Question

Help with this integral="TexFormula1" title=" \int\limits \frac{dx}{x^{2}-4x-13}" alt=" \int\limits \frac{dx}{x^{2}-4x-13}" align="absmiddle" class="latex-formula">

Answer 1

$x^2-4x-13=(x-2)^2-17$

$x-2=\sqrt{17}\sec y$
$\mathrm dx=\sqrt{17}\sec y\tan y\,\mathrm dy$

$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=\int\frac{\sqrt{17}\sec y\tan y}{(\sqrt{17}\sec y)^2-17}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\sec^2y-1}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\tan^2y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y}{\tan y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\frac1{\cos y}}{\frac{\sin y}{\cos y}}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\csc y\,\mathrm dy$
$=-\dfrac1{\sqrt{17}}\ln|\csc y+\cot y|+C$

$\sec y=\dfrac{x-2}{\sqrt{17}}\iff y=\sec^{-1}\dfrac{x-2}{\sqrt{17}}$
$\implies\csc y=\dfrac{x-2}{\sqrt{(x-2)^2-17}}=\dfrac{x-2}{\sqrt{x^2-4x-13}}$
$\implies\cot y=\dfrac{\sqrt{17}}{\sqrt{(x-2)^2-17}}=\dfrac{\sqrt{17}}{\sqrt{x^2-4x-13}}$

$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=-\dfrac1{\sqrt{17}}\ln\left|\frac{x-2+\sqrt{17}}{\sqrt{x^2-4x-13}}\right|+C$