What is the integral of tanh(2x)ln(cosh(2x)) ?

The integral of tanh(2x)ln(cosh(2x)) is 1/4 ln^2(cosh(2x))+C

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integral of tanh(2x)ln(cosh(2x))

\int tanh (2 x) ln (cosh (2 x)) d x

\frac{1}{4} ln^{2} (cosh (2 x)) + C

Solution steps

\int tanh (2 x) ln (cosh (2 x)) d x

Apply u-substitution

: \int \frac{u}{2} d u

= \int \frac{u}{2} 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 u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

u = ln (cosh (2 x))

= \frac{1}{2} \cdot \frac{ln ^{2} ( cosh ( 2 x ) )}{2}

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( cosh ( 2 x ) )}{2} : \frac{1}{4} ln^{2} (cosh (2 x))

= \frac{1}{4} ln^{2} (cosh (2 x))

Add a constant to the solution

= \frac{1}{4} ln^{2} (cosh (2 x)) + C

cos (x), x^{2} + 2, [1, 2]

\int \frac{5}{x ( x ^{2} + 2 ) ^{2}} d x

\frac{s}{s ^{2} + s}

\frac{\partial}{\partial x} (4 x y e^{x y^{2}})

\int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x

What is the integral of tanh(2x)ln(cosh(2x)) ?
The integral of tanh(2x)ln(cosh(2x)) is 1/4 ln^2(cosh(2x))+C