Help with this integral \int\limits (dx)/(x^(2)-4x-13)

Question

Help with this integral


`\int\limits (dx)/(x^(2)-4x-13)`

Answer 1

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


`x-2=√(17)\sec y`

`\mathrm dx=√(17)\sec y\tan y\,\mathrm dy`


`\displaystyle\int(\mathrm dx)/(x^2-4x-13)=\int(√(17)\sec y\tan y)/((√(17)\sec y)^2-17)\,\mathrm dy`

`=\displaystyle\frac1{√(17)}\int(\sec y\tan y)/(\sec^2y-1)\,\mathrm dy`

`=\displaystyle\frac1{√(17)}\int(\sec y\tan y)/(\tan^2y)\,\mathrm dy`

`=\displaystyle\frac1{√(17)}\int(\sec y)/(\tan y)\,\mathrm dy`

`=\displaystyle\frac1{√(17)}\int\frac{\frac1{\cos y}}{(\sin y)/(\cos y)}\,\mathrm dy`

`=\displaystyle\frac1{√(17)}\int\csc y\,\mathrm dy`

`=-\frac1{√(17)}\ln|\csc y+\cot y|+C`


`\sec y=(x-2)/(√(17))\iff y=\sec^(-1)(x-2)/(√(17))`

`\implies\csc y=(x-2)/(√((x-2)^2-17))=(x-2)/(√(x^2-4x-13))`

`\implies\cot y=(√(17))/(√((x-2)^2-17))=(√(17))/(√(x^2-4x-13))`


`\displaystyle\int(\mathrm dx)/(x^2-4x-13)=-\frac1{√(17)}\ln\left|(x-2+√(17))/(√(x^2-4x-13))\right|+C`