What is the integral of (e^x)/(sqrt(16-e^{2x))} ?

The integral of (e^x)/(sqrt(16-e^{2x))} is arcsin(1/4 e^x)+C

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integral of (e^x)/(sqrt(16-e^{2x))}

\int \frac{e ^{x}}{16 - e ^{2 x}} d x

arcsin (\frac{1}{4} e^{x}) + C

Solution steps

\int \frac{e ^{x}}{16 - e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{16 - u ^{2}} d u

= \int \frac{1}{16 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{4} e^{x})

Simplify

= arcsin (\frac{1}{4} e^{x})

Add a constant to the solution

= arcsin (\frac{1}{4} e^{x}) + C

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What is the integral of (e^x)/(sqrt(16-e^{2x))} ?
The integral of (e^x)/(sqrt(16-e^{2x))} is arcsin(1/4 e^x)+C