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

The integral from 0 to 2 of ln(x^2+1)x is 1/2 (5ln(5)-4)

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

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

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

Decimal

2.02359 \dots

Solution steps

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

Apply u-substitution

= \int_{1}^{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 \int_{1}^{5} ln (u) d u

Apply Integration By Parts

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

\int 1 d u = u

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

Compute the boundaries:

5 ln (5) - 4

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

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What is the integral from 0 to 2 of ln(x^2+1)x ?
The integral from 0 to 2 of ln(x^2+1)x is 1/2 (5ln(5)-4)