What is the integral of 16csch^2(x)coth(x) ?

The integral of 16csch^2(x)coth(x) is -8csch^2(x)+C

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integral of 16csch^2(x)coth(x)

\int 16 csch^{2} (x) coth (x) d x

- 8 csch^{2} (x) + C

Solution steps

\int 16 csch^{2} (x) coth (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 16 \cdot \int csch^{2} (x) coth (x) d x

Apply u-substitution

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

= 16 \cdot \int - \frac{1}{2} d u

Integral of a constant:

\int a d x = a x

= 16 (- \frac{1}{2}) u

Substitute back

u = csch^{2} (x)

= 16 (- \frac{1}{2}) csch^{2} (x)

Simplify

16 (- \frac{1}{2}) csch^{2} (x) : - 8 csch^{2} (x)

= - 8 csch^{2} (x)

Add a constant to the solution

= - 8 csch^{2} (x) + C

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What is the integral of 16csch^2(x)coth(x) ?
The integral of 16csch^2(x)coth(x) is -8csch^2(x)+C