Question

Evaluate : $\int \frac{dx}{(x^2+3)\sqrt{x-1}}dx$

Answer 1

$\int \frac{1}{\sqrt{x-1}(x^2+3)}dx$

substitute $u=\sqrt{x-1}\to dx=2\sqrt{x-1}du$

$=2\int \frac{1}{(u^2+1{)}^2+3}du$

$=2\int \frac{1}{(u^2-\sqrt{2}u+2)(u^2+\sqrt{2}u+2)}du$

$=2\int (\frac{u+\sqrt{2}}{2^{\frac{5}{2}}(u^2+\sqrt{2}u+2)}-\frac{u-\sqrt{2}}{2^{\frac{5}{2}}(u^2-\sqrt{2}u+2)})du$

$=2\left(\frac{1}{2^{\frac{5}{2}}}\right)[\int \frac{u+\sqrt{2}}{u^2+\sqrt{2}u+2}-\int \frac{u-\sqrt{2}}{u^2-\sqrt{2}u+2}]$

$=\frac{\ln (u^2+\sqrt{2}u+2)}{2^{\frac{7}{2}}}-\frac{\ln (u^2-\sqrt{2}u+2)}{2^{\frac{7}{2}}}+\frac{{\tan }^{-1}\left(\frac{\sqrt{2}u+1}{\sqrt{3}}\right)}{2^{\frac{5}{2}}\sqrt{3}}+\frac{{\tan }^{-1}\left(\frac{\sqrt{2}u-1}{\sqrt{3}}\right)}{2^{\frac{5}{2}}\sqrt{3}}$

$=\frac{\ln (x+\sqrt{2}\sqrt{x-1}+1-\ln (|x-\sqrt{2}\sqrt{x-1}+1|)}{2^{\frac{5}{2}}}+\frac{{\tan }^{-1}\left(\frac{\sqrt{2}\sqrt{x-1}+1}{\sqrt{3}}\right)+{\tan }^{-1}\left(\frac{\sqrt{2}\sqrt{x-1}-1}{\sqrt{3}}\right)}{2^{\frac{3}{2}}\sqrt{3}}+C$