Question

Evaluate :
$\int \frac{x-5}{\sqrt{x^2+6x+7}}dx.$

Answer 1

$I=\int \frac{x-5}{\sqrt{x^2+6x+7}}dx=\int \frac{x+3-8}{\sqrt{x^2+6x+7}}dx=\int \frac{x+3}{\sqrt{x^2+6x+7}}dx-\int \frac{8}{\sqrt{x^2+6x+7}}dx$

$A=\int \frac{x+3}{\sqrt{x^2+6x+7}}dx$

$x^2+6x+7=t^2$

$(2x+6)dx=2tdt\Rightarrow (x+3)dx=tdt$

$A=\int \frac{tdt}{t}\Rightarrow A=t+C\Rightarrow A=\sqrt{x^2+6x+7}+C$

$B=\int \frac{8}{\sqrt{x^2+6x+7}}dx$

$x+3=p\Rightarrow dx=dp$

$B=\int \frac{8dp}{\sqrt{p^2-{\left(\sqrt{2}\right)}^2}}$

$p=\sqrt{2}\sec \theta \Rightarrow dp=\sqrt{2}\sec \theta -\tan \theta d\theta$

$B=\int \frac{8\sqrt{2}\sec \theta -\tan \theta d\theta }{\sqrt{2{\sec }^2\theta -2}}=\int \frac{8\sec \theta -\tan \theta d\theta }{\tan \theta }=\int 8\sec \theta d\theta =8\ln |\sec \theta +\tan \theta |=8\ln |\frac{p}{\sqrt{2}}+\sqrt{\frac{p^2}{2}-1}|=8\ln |\frac{x+3}{\sqrt{2}}+\sqrt{\frac{{(x+3)}^2-2}{\sqrt{2}}}|$

$\therefore$ $I=\sqrt{x^2+6x+7}+8\ln \left|\frac{x+3+\sqrt{x^2+6x+7}}{\sqrt{2}}\right|+C$