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

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

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

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

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

Decimal

0.34657 \dots

Solution steps

\int_{0}^{1} \frac{x}{1 + x ^{2}} 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}^{5} x d x

\int_{- a}^{a} 1 d x

\int_{1}^{2} (32 x^{3} + \frac{6}{x ^{2}}) d x

\int_{1}^{3} (8 x - 1) e^{4 x^{2} - x} d x

\int_{0}^{5} 1 + 4 x^{2} d x

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