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

The integral of 1/(sqrt(e^{2x)+9)} is -1/6 (ln|(sqrt(e^{2x)+9})/3+1|-ln|(sqrt(e^{2x)+9})/3-1|)+C

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integral of 1/(sqrt(e^{2x)+9)}

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

- \frac{1}{6} (ln \frac{e ^{2 x} + 9}{3} + 1 - ln \frac{e ^{2 x} + 9}{3} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{v v + 9} d v

= \frac{1}{2} \cdot \int \frac{1}{v v + 9} d v

Apply u-substitution

: \int \frac{2}{w ^{2} - 9} d w

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

Take the constant out:

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{3 ( t ^{2} - 1 )} d t

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{t ^{2} - 1} d t

Factor

t^{2} - 1 : - (- t^{2} + 1)

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{- ( - t ^{2} + 1 )} d t

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} (- \int \frac{1}{- t ^{2} + 1} d t)

Use the common integral:

\int \frac{1}{- t ^{2} + 1} d t = \frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} (- (\frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2}))

Substitute back

= \frac{1}{2} \cdot 2 \cdot \frac{1}{3} - \frac{ln \frac{e ^{2 x} + 9}{3} + 1}{2} - \frac{ln \frac{e ^{2 x} + 9}{3} - 1}{2}

Simplify

\frac{1}{2} \cdot 2 \cdot \frac{1}{3} - \frac{ln \frac{e ^{2 x} + 9}{3} + 1}{2} - \frac{ln \frac{e ^{2 x} + 9}{3} - 1}{2} : - \frac{1}{6} (ln \frac{e ^{2 x} + 9}{3} + 1 - ln \frac{e ^{2 x} + 9}{3} - 1)

= - \frac{1}{6} (ln \frac{e ^{2 x} + 9}{3} + 1 - ln \frac{e ^{2 x} + 9}{3} - 1)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{e ^{2 x} + 9}{3} + 1 - ln \frac{e ^{2 x} + 9}{3} - 1) + C

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What is the integral of 1/(sqrt(e^{2x)+9)} ?
The integral of 1/(sqrt(e^{2x)+9)} is -1/6 (ln|(sqrt(e^{2x)+9})/3+1|-ln|(sqrt(e^{2x)+9})/3-1|)+C