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

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

\int x 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

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 = 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

\frac{d y}{d x} = (\frac{y}{x})

\frac{d}{d x} (x^{4} sin (x) sin (3 x))

\frac{d}{d x} (3 (1 - x)^{- 2})

y^{^{'}} = \frac{sin ( x )}{y}

\int x sin (- 4 x) d x

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