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

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

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

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

2 ln (2) - 1

Decimal

0.38629 \dots

Solution steps

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

Simplify

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

= \int_{0}^{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

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

Apply u-substitution

: \int_{1}^{2} \frac{ln ( u )}{2} d u

= 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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Simplify

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

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

Compute the boundaries:

2 ln (2) - 1

= 2 ln (2) - 1

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