What is the integral of xcsch^2((x^2)/2) ?

The integral of xcsch^2((x^2)/2) is -coth((x^2)/2)+C

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integral of xcsch^2((x^2)/2)

\int x csch^{2} (\frac{x ^{2}}{2}) d x

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

Solution steps

\int x csch^{2} (\frac{x ^{2}}{2}) d x

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

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

Expand

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

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

Apply u-substitution

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

Apply u-substitution

= \frac{1}{2} \cdot \int csch^{2} (v) \cdot 2 d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int csch^{2} (v) d v

Use the common integral:

\int csch^{2} (v) d v = - coth (v)

= \frac{1}{2} \cdot 2 (- coth (v))

Substitute back

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

Simplify

\frac{1}{2} \cdot 2 (- coth (\frac{x ^{2}}{2})) : - coth (\frac{x ^{2}}{2})

= - coth (\frac{x ^{2}}{2})

Add a constant to the solution

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

d er i v a t i v e \frac{( x ^{2} - 1 )}{( x ^{2} + x + 1 )}

d er i v a t i v e f (x) = (8 (x^{3} - 6 x^{2} + 12 x))

\frac{\partial}{\partial x} (\frac{cos ( x ^{2} \cdot y )}{z ^{3}})

im pl i c i t \frac{d y}{d x}, 5 x y^{5} + 3 x y = 24

d er i v a t i v e arctan (\frac{x}{6})

What is the integral of xcsch^2((x^2)/2) ?
The integral of xcsch^2((x^2)/2) is -coth((x^2)/2)+C