What is the integral from 0 to 1 of xln(sqrt(x^2+1)) ?

The integral from 0 to 1 of xln(sqrt(x^2+1)) is 1/4 (2ln(2)-1)

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integral from 0 to 1 of xln(sqrt(x^2+1))

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

\frac{1}{4} (2 ln (2) - 1)

Decimal

0.09657 \dots

Solution steps

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

Simplify

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

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

Apply u-substitution

= \frac{1}{2} \cdot \int_{1}^{2} \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_{1}^{2} ln (u) d u

Apply Integration By Parts

= \frac{1}{2} \cdot \frac{1}{2} [u ln (u) - \int 1 d u]_{1}^{2}

\int 1 d u = u

= \frac{1}{2} \cdot \frac{1}{2} [u ln (u) - u]_{1}^{2}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [u ln (u) - u]_{1}^{2} : \frac{1}{4} [u ln (u) - u]_{1}^{2}

= \frac{1}{4} [u ln (u) - u]_{1}^{2}

Compute the boundaries:

2 ln (2) - 1

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

What is the integral from 0 to 1 of xln(sqrt(x^2+1)) ?
The integral from 0 to 1 of xln(sqrt(x^2+1)) is 1/4 (2ln(2)-1)