What is going on with this integral ∫ d x e 2 x − 9

Question

What is going on with this integral   ∫            d      x                      e                  2          x                    −      9

Myla Pierce · Accepted Answer

Rewrite the integrand by multiplying and dividing by

e^{x}

\int \frac{e^{x} d x}{e^{x} \sqrt{e^{x} - 9}} = \int \frac{d (e^{x})}{e^{x} \sqrt{e^{x} - 9}} = \frac{1}{3} \sec^{- 1} (\frac{e^{x}}{3}) + C

By drawing a triangle we can see that

\sec 3 θ = \frac{e^{x}}{3} ⟹ \tan 3 θ = \frac{\sqrt{e^{2 x} - 9}}{3}

Thus we also obtain your arctan solution, which is equivalent to the arcsec solution

\frac{1}{3} \tan^{- 1} (\frac{\sqrt{e^{2 x} - 9}}{3}) + C

and similarly, we can obtain the arcsin solution

\sin 3 θ = \frac{e^{x}}{\sqrt{e^{2 x} - 9}} ⟹ \frac{1}{3} \arcsin (\frac{1}{\sqrt{1 - 9 e^{- x}}}) + C

What is going on with this integral ∫ d x </mrow> <m