Question

integral of 1 sqrt x^2+4x+8

Answer 1

\[\int\limits_{}^{}\sqrt{x^2+4x+8} dx\] ?
is that right?

Answer 2

\[\int\limits_{}^{}\sqrt{(x^2+4x+4)+4}dx=\int\limits_{}^{}\sqrt{(x+2)^2+4}dx\]
let \[\tan \theta=\frac{x+2}{2}\]
\[\sec^2 \theta d \theta=\frac{1}{2} dx\]
\[\int\limits_{}^{}\sqrt{4*(\frac{x+2}{2})^2+4}dx=\int\limits_{}^{}\sqrt{4}\sqrt{\tan^2\theta+1}*2\sec^2\theta d \theta\]
\[4\int\limits_{}^{}\sec^3\theta d \theta\]
this should help