What is the integral of 1/(sinh^2(x)+cosh^2(x)) ?

The integral of 1/(sinh^2(x)+cosh^2(x)) is 1/2 arctan(sinh(2x))+C

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integral of 1/(sinh^2(x)+cosh^2(x))

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

\frac{1}{2} arctan (sinh (2 x)) + C

Solution steps

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

Rewrite using trig identities

= \int sech (2 x) d x

Apply u-substitution

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

= \int sech (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 (u) d u

Use the common integral:

\int sech (u) d u = arctan (sinh (u))

= \frac{1}{2} arctan (sinh (u))

Substitute back

u = 2 x

= \frac{1}{2} arctan (sinh (2 x))

Add a constant to the solution

= \frac{1}{2} arctan (sinh (2 x)) + C

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What is the integral of 1/(sinh^2(x)+cosh^2(x)) ?
The integral of 1/(sinh^2(x)+cosh^2(x)) is 1/2 arctan(sinh(2x))+C