What is the integral of sech^2(x)tanh(x) ?

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

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integral of sech^2(x)tanh(x)

\int sech^{2} (x) tanh (x) d x

- \frac{1}{2} sech^{2} (x) + C

Solution steps

\int sech^{2} (x) tanh (x) d x

Apply u-substitution

: \int - \frac{1}{2} d u

= \int - \frac{1}{2} d u

Integral of a constant:

\int a d x = a x

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

Substitute back

u = sech^{2} (x)

= (- \frac{1}{2}) sech^{2} (x)

Simplify

= - \frac{1}{2} sech^{2} (x)

Add a constant to the solution

= - \frac{1}{2} sech^{2} (x) + C

x \to 3 - lim (\frac{2}{x ^{2} - 9})

e^{9 - x}

y^{^{''}} + 25 y = cos (6 t)

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

\int \frac{5 x ^{2}}{( x - 1 ) ( x ^{2} + 9 )} d x

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