Question

Find $\int \sqrt{x^2+2x+5}.dx$

Answer 1

To find: $\int \sqrt{x^2+2x+5}.dx$

=$\int \sqrt{x^2+2x+1+4}.dx$

=$\int \sqrt{(x+1{)}^2+4}.dx$

Let $t=x+1$

$\Rightarrow dt=dx$

Substituting,

$=\int \sqrt{t^2+4}.dt$

$=\int \sqrt{t^2+(2{)}^2}.dt$

$[\text{Using}\int \sqrt{(x^2+a^2)}.dx=\frac{1}{2}x\sqrt{(x^2+a^2)}+\frac{a^2}{2}\log |x+\sqrt{(x^2+a^2)}|]$

$=\frac{t}{2}\sqrt{t^2+4}+\frac{4}{2}\log |x+\sqrt{(t^2+4)}|+C$

$=\frac{t}{2}\sqrt{t^2+4}+2\log |x+\sqrt{(t^2+4)}|+C$

Putting the value of $t=x+1$

$=\frac{x+1}{2}\sqrt{(x+1{)}^2+4}+2\log |x+1+\sqrt{(x+1{)}^2+4}|+C$

$=\frac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log |x+1+\sqrt{x^2+2x+5}|+C$

Where $C$ is constant of integration.