What is the integral of e^{3z}tanh(e^{3z})ln(cosh(e^{3z})) ?

The integral of e^{3z}tanh(e^{3z})ln(cosh(e^{3z})) is 1/6 ln^2(cosh(e^{3z}))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{3z}tanh(e^{3z})ln(cosh(e^{3z}))

\int e^{3 z} tanh (e^{3 z}) ln (cosh (e^{3 z})) d z

\frac{1}{6} ln^{2} (cosh (e^{3 z})) + C

Solution steps

\int e^{3 z} tanh (e^{3 z}) ln (cosh (e^{3 z})) d z

Apply u-substitution

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

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

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{2}}{2}

Substitute back

u = ln (cosh (e^{3 z}))

= \frac{1}{3} \cdot \frac{ln ^{2} ( cosh ( e ^{3 z} ) )}{2}

Simplify

\frac{1}{3} \cdot \frac{ln ^{2} ( cosh ( e ^{3 z} ) )}{2} : \frac{1}{6} ln^{2} (cosh (e^{3 z}))

= \frac{1}{6} ln^{2} (cosh (e^{3 z}))

Add a constant to the solution

= \frac{1}{6} ln^{2} (cosh (e^{3 z})) + C

\int \frac{cos ^{3} ( x )}{sin ^{5} ( x )} d x

\int \frac{( 2 + x ) ^{6}}{x} d x

\int \frac{( 2 - 4 x ) ^{2}}{x ^{2}} d x

\int \frac{1}{2 + 4 x ^{4}} (16 x^{3}) d x

\int \frac{( x ^{2} - 1 ) ^{3}}{x} d x

What is the integral of e^{3z}tanh(e^{3z})ln(cosh(e^{3z})) ?
The integral of e^{3z}tanh(e^{3z})ln(cosh(e^{3z})) is 1/6 ln^2(cosh(e^{3z}))+C