What is the integral of (cosh(3x))/(1+sinh(3x)) ?

The integral of (cosh(3x))/(1+sinh(3x)) is 1/3 ln|1+sinh(3x)|+C

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integral of (cosh(3x))/(1+sinh(3x))

\int \frac{cosh ( 3 x )}{1 + sinh ( 3 x )} d x

\frac{1}{3} ln ∣ 1 + sinh (3 x) ∣ + C

Solution steps

\int \frac{cosh ( 3 x )}{1 + sinh ( 3 x )} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + sinh (3 x)

= \frac{1}{3} ln ∣ 1 + sinh (3 x) ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 1 + sinh (3 x) ∣ + C

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What is the integral of (cosh(3x))/(1+sinh(3x)) ?
The integral of (cosh(3x))/(1+sinh(3x)) is 1/3 ln|1+sinh(3x)|+C