What is the integral from 0 to 6 of x/(x^2+1) ?

The integral from 0 to 6 of x/(x^2+1) is 1/2 ln(37)

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integral from 0 to 6 of x/(x^2+1)

\int_{0}^{6} \frac{x}{x ^{2} + 1} d x

\frac{1}{2} ln (37)

Solution steps

\int_{0}^{6} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

: \int_{1}^{37} \frac{1}{2 u} d u

= \int_{1}^{37} \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}^{37} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{37}

Compute the boundaries:

ln (37)

= \frac{1}{2} ln (37)

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What is the integral from 0 to 6 of x/(x^2+1) ?
The integral from 0 to 6 of x/(x^2+1) is 1/2 ln(37)