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

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

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

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

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

Solution steps

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

Simplify

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

= \int \frac{1}{2} 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

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

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{4} (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(sqrt(1+x^2)) ?
The integral of x*ln(sqrt(1+x^2)) is 1/4 (ln(1+x^2)+x^2ln(1+x^2)-1-x^2)+C