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

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

\int x \cdot 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

\int_{0}^{1} ln (2 x + 1) d x

f (x) = - 4 x^{2} - 3 + 2, at x = - 1

\int_{0}^{1} \frac{x + 1}{x ^{2} + 1} d x

\int \frac{e ^{\frac{1}{x ^{8}}}}{x ^{9}} d x

x \to - 3 lim (\frac{x - 3}{9 x})

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