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

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

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

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

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

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{ln ( u )}{2} d u

= \frac{1}{2} \cdot \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 \frac{1}{2} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{2} \cdot \frac{1}{2} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2} \cdot \frac{1}{2} (u ln (u) - u)

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

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