What is the integral of 1/(e^xsqrt(e^{2x)-9)} ?

The integral of 1/(e^xsqrt(e^{2x)-9)} is (sqrt(e^{2x)-9})/(9e^x)+C

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integral of 1/(e^xsqrt(e^{2x)-9)}

\int \frac{1}{e ^{x} e ^{2 x} - 9} d x

\frac{e ^{2 x} - 9}{9 e ^{x}} + C

Solution steps

\int \frac{1}{e ^{x} e ^{2 x} - 9} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{9 sec ( v )} d v

= \int \frac{1}{9 sec ( v )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= \frac{1}{9} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= \frac{1}{9} sin (v)

Substitute back

= \frac{1}{9} sin (arcsec (\frac{1}{3} e^{x}))

Simplify

\frac{1}{9} sin (arcsec (\frac{1}{3} e^{x})) : \frac{e ^{2 x} - 9}{9 e ^{x}}

= \frac{e ^{2 x} - 9}{9 e ^{x}}

Add a constant to the solution

= \frac{e ^{2 x} - 9}{9 e ^{x}} + C

sin (x), 0, 2 π

\int x^{15} d x

\frac{d}{d x} ((7 x^{2} - 12 x) e^{x})

\int (x^{2} - 9)^{5} (2 x) d x

\frac{\partial}{\partial y} (4 x^{3} - 5 x^{2} + 8 x)

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