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

The integral of-xln(2x^2+1) is -1/2 x^2ln(2x^2+1)+(x^2)/2-1/4 ln|2x^2+1|+C

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integral of-xln(2x^2+1)

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

- \frac{1}{2} x^{2} ln (2 x^{2} + 1) + \frac{x ^{2}}{2} - \frac{1}{4} ln 2 x^{2} + 1 + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: \frac{1}{2} x^{2} ln (2 x^{2} + 1) - \int \frac{2 x ^{3}}{2 x ^{2} + 1} d x

= - (\frac{1}{2} x^{2} ln (2 x^{2} + 1) - \int \frac{2 x ^{3}}{2 x ^{2} + 1} d x)

\int \frac{2 x ^{3}}{2 x ^{2} + 1} d x = 2 (\frac{x ^{2}}{4} - \frac{1}{8} ln 2 x^{2} + 1)

= - (\frac{1}{2} x^{2} ln (2 x^{2} + 1) - 2 (\frac{x ^{2}}{4} - \frac{1}{8} ln 2 x^{2} + 1))

Simplify

- (\frac{1}{2} x^{2} ln (2 x^{2} + 1) - 2 (\frac{x ^{2}}{4} - \frac{1}{8} ln 2 x^{2} + 1)) : - \frac{1}{2} x^{2} ln (2 x^{2} + 1) + \frac{x ^{2}}{2} - \frac{1}{4} ln 2 x^{2} + 1

= - \frac{1}{2} x^{2} ln (2 x^{2} + 1) + \frac{x ^{2}}{2} - \frac{1}{4} ln 2 x^{2} + 1

Add a constant to the solution

= - \frac{1}{2} x^{2} ln (2 x^{2} + 1) + \frac{x ^{2}}{2} - \frac{1}{4} ln 2 x^{2} + 1 + C

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What is the integral of-xln(2x^2+1) ?
The integral of-xln(2x^2+1) is -1/2 x^2ln(2x^2+1)+(x^2)/2-1/4 ln|2x^2+1|+C