Question

Integrate the function: $\sqrt{x^2+4x+6}$

Answer 1

Let $I=\int \sqrt{x^2+4x+6} \,dx$

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

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

=$\int \sqrt{(x+2)^2+(\sqrt2)^2}dx$

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

∴$I = \frac{(x+2)}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\log \mid(x+2)+\sqrt{x^2+4x+6}\mid+C$

=$\frac{(x+2)}{2}\sqrt{x^2+4x+6}+\log \mid (X+2)+\sqrt{x^2+4x+6}\mid+C$