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

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

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

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

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

Decimal

0.34657 \dots

Solution steps

\int_{0}^{1} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{2 u} d u

= \int_{1}^{2} \frac{1}{2 u} 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} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

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

\int_{1}^{2} \frac{2}{x ^{2}} - 3 x^{2} - 1 d x

\int_{0}^{π} cos (x) sin^{4} (x) d x

\int_{2}^{4} x^{2} - 4 d x

\int_{0}^{1} \frac{1}{1 + y ^{2}} d y

\int_{0}^{2} \frac{1}{( 1 - 2 x ) ^{2}} d x

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