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

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

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integral of [xln(x^2+1)]

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

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

Solution steps

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

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

Expand

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

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

Add a constant to the solution

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

y^{^{''}} + y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

x y y^{^{'}} = x^{2} e^{\frac{y}{x}} + y^{2}

\frac{d}{d x} (a^{2} x)

(x + 3) (\frac{d y}{d x}) + y = ln (x)

\frac{d ^{2}}{d x ^{2}} ((9 x^{2} + 6)^{6})

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