What is the integral of (e^x)/(sqrt(e^{2x)+9)} ?

The integral of (e^x)/(sqrt(e^{2x)+9)} is ln(1/3 |e^x+sqrt(9+e^{2x)}|)+C

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integral of (e^x)/(sqrt(e^{2x)+9)}

\int \frac{e ^{x}}{e ^{2 x} + 9} d x

ln (\frac{1}{3} e^{x} + 9 + e^{2 x}) + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{3} e^{x})) + sec (arctan (\frac{1}{3} e^{x}))

Simplify

ln tan (arctan (\frac{1}{3} e^{x})) + sec (arctan (\frac{1}{3} e^{x})) : ln (\frac{1}{3} e^{x} + 9 + e^{2 x})

= ln (\frac{1}{3} e^{x} + 9 + e^{2 x})

Add a constant to the solution

= ln (\frac{1}{3} e^{x} + 9 + e^{2 x}) + C

y = 5 x^{2} - 12 x - \frac{8}{x ^{2}}

- \frac{1}{2 x ^{\frac{3}{2}}}

\int 2 tanh (x) d x

ln (7)

\int x (6 + 5 x) d x

What is the integral of (e^x)/(sqrt(e^{2x)+9)} ?
The integral of (e^x)/(sqrt(e^{2x)+9)} is ln(1/3 |e^x+sqrt(9+e^{2x)}|)+C