Question

Integrate the function: $\sqrt{x^2+4x+1}$

Answer 1

Let $I=\int \sqrt{x^2+4x+1}\,dx$

=$\int \sqrt{(x^2+4x+4)-3}dx$

=$\int \sqrt{(x+2)^2-(\sqrt3)^2}dx$

It is known that,$\int \sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log \mid x+\sqrt{x^2-a^2}\mid+C$

∴=$I=\frac{(x+2)}{2}\sqrt{x^2+4x+1-}\frac{3}{2}\log\mid (x+2)+\sqrt{x^2+4x+1}\mid+C$