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

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

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

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

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

Decimal

0.34657 \dots

Solution steps

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

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \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})

Simplify

= \frac{1}{2} [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

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

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

\int_{0}^{1} a x^{2} d x

\int_{0}^{2} \frac{8 x}{2 x ^{2} + 1} d x

\int_{0}^{2} - 4 d x

\int_{1}^{8} x^{2} - 4 x - 12 d x

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