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

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

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integral of-x*ln(1+x^2)

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

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

Solution steps

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

Take the constant out:

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

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

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

Expand

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

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

Add a constant to the solution

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

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