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

The integral of xln(x^2+2) is 1/2 x^2ln(x^2+2)+ln(x^2+2)-1/2 x^2-1+C

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integral of xln(x^2+2)

\int x ln (x^{2} + 2) d x

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

Solution steps

\int x ln (x^{2} + 2) d x

Apply u-substitution

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

= \int \frac{ln ( 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 ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

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

\int 1 d u = u

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

Substitute back

u = x^{2} + 2

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

Simplify

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

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

Add a constant to the solution

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

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