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

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

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

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

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

Decimal Notation

0.19314 \dots

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Compute the boundaries:

2 ln (2) - 1

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

Simplify

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

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