What is the integral of sech^2(2x-1/8) ?

The integral of sech^2(2x-1/8) is 1/2 tanh(2x-1/8)+C

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integral of sech^2(2x-1/8)

\int sech^{2} (2 x - \frac{1}{8}) d x

\frac{1}{2} tanh (2 x - \frac{1}{8}) + C

Solution steps

\int sech^{2} (2 x - \frac{1}{8}) d x

Apply u-substitution

: \int sech^{2} (u) \frac{1}{2} d u

= \int sech^{2} (u) \frac{1}{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 sech^{2} (u) d u

Use the common integral:

\int sech^{2} (u) d u = tanh (u)

= \frac{1}{2} tanh (u)

Substitute back

u = 2 x - \frac{1}{8}

= \frac{1}{2} tanh (2 x - \frac{1}{8})

Add a constant to the solution

= \frac{1}{2} tanh (2 x - \frac{1}{8}) + C

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What is the integral of sech^2(2x-1/8) ?
The integral of sech^2(2x-1/8) is 1/2 tanh(2x-1/8)+C