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

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

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integral of 1/(sqrt(9-e^{2t))}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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