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

The integral from 1 to 2 of xln(sqrt(1+x^2)) is 1/4 (5ln(5)-3-2ln(2))

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

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

\frac{1}{4} (5 ln (5) - 3 - 2 ln (2))

Decimal

0.91522 \dots

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int_{2}^{5} \frac{ln ( u )}{2} d u

= \frac{1}{2} \cdot \int_{2}^{5} \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_{2}^{5} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{2}^{5}

= \frac{1}{2} \cdot \frac{1}{2} [u ln (u) - \int 1 d u]_{2}^{5}

\int 1 d u = u

= \frac{1}{2} \cdot \frac{1}{2} [u ln (u) - u]_{2}^{5}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [u ln (u) - u]_{2}^{5} : \frac{1}{4} [u ln (u) - u]_{2}^{5}

= \frac{1}{4} [u ln (u) - u]_{2}^{5}

Compute the boundaries:

5 ln (5) - 3 - 2 ln (2)

= \frac{1}{4} (5 ln (5) - 3 - 2 ln (2))

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What is the integral from 1 to 2 of xln(sqrt(1+x^2)) ?
The integral from 1 to 2 of xln(sqrt(1+x^2)) is 1/4 (5ln(5)-3-2ln(2))