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

The integral of xtan(x^2+1) is -1/2 ln|cos(x^2+1)|+C

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integral of xtan(x^2+1)

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{tan ( u + 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 tan (u + 1) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{sin ( u + 1 )}{cos ( u + 1 )} d u

Apply u-substitution

: \int - \frac{1}{v} d v

= \frac{1}{2} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= \frac{1}{2} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} (- ln ∣ v ∣)

Substitute back

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

Simplify

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

Add a constant to the solution

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

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What is the integral of xtan(x^2+1) ?
The integral of xtan(x^2+1) is -1/2 ln|cos(x^2+1)|+C