What is the integral of xcosh(x^2+1) ?

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

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integral of xcosh(x^2+1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{cosh ( u )}{2} d u

= \int \frac{cosh ( u )}{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 cosh (u) d u

Use the common integral:

\int cosh (u) d u = sinh (u)

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

Substitute back

u = x^{2} + 1

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

Add a constant to the solution

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

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