Question

Evaluate:
$\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$.

Answer 1

$I=\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=5\int \frac{x+\frac{3}{5}}{\sqrt{x^2+4x+10}}dx$

$=\frac{5}{2}\int \frac{2x+\frac{6}{5}}{\sqrt{x^2+4x+10}}dx$

$=\frac{5}{2}\int \frac{2x+4-4+\frac{6}{5}}{\sqrt{x^2+4x+10}}dx$

$=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx+\frac{5}{2}\int \frac{\frac{-14}{5}}{\sqrt{x^2+4x+10}}dx$

$=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}dx$

Let $I_1=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$ and $I_2=7\int \frac{dx}{\sqrt{x^2+4x+10}}dx$

$I_1=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx$

Let $x^2+4x+10=t$

$(2x+4)dx=dt$

Thus, $I_1=\frac{5}{2}\int \frac{1}{\sqrt{t}}dt$

$I_1=\frac{5}{2}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C_1$

$I_1=5\sqrt{t}+C_1$

$I_1=5\sqrt{x^2+4x+10}+C_1$

Now, $I_2=7\int \frac{dx}{\sqrt{x^2+4x+10}}dx$

$I_2=7\int \frac{dx}{\sqrt{(x+2{)}^2+(\sqrt{6}{)}^2}}dx$

$I_2=7[log|x+2+\sqrt{x^2+4x+10}|]+C_2$

Therefore, $I=5\sqrt{x^2+4x+10}-7log|x+2+\sqrt{x^2+4x+10}|+C$