What is the integral of sech^8(x)tanh(x) ?

The integral of sech^8(x)tanh(x) is -1/8 sech^8(x)+C

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integral of sech^8(x)tanh(x)

\int sech^{8} (x) tanh (x) d x

- \frac{1}{8} sech^{8} (x) + C

Solution steps

\int sech^{8} (x) tanh (x) d x

Apply u-substitution

: \int - \frac{1}{8} d u

= \int - \frac{1}{8} d u

Integral of a constant:

\int a d x = a x

= (- \frac{1}{8}) u

Substitute back

u = sech^{8} (x)

= (- \frac{1}{8}) sech^{8} (x)

Simplify

= - \frac{1}{8} sech^{8} (x)

Add a constant to the solution

= - \frac{1}{8} sech^{8} (x) + C

\frac{\partial}{\partial x} (x^{2} y + y^{3} - 27 y)

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

x \to 2 lim (x^{2} + 3 x + 4)

\int e^{- a x} d x

\frac{d y}{d x} = \frac{8}{7} y

What is the integral of sech^8(x)tanh(x) ?
The integral of sech^8(x)tanh(x) is -1/8 sech^8(x)+C