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

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

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integral of xln(2+x^2)

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

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

Solution steps

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

Apply u-substitution

= \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

= \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 = 2 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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