What is the integral from 1 to 4 of x/(1+x^2) ?

The integral from 1 to 4 of x/(1+x^2) is 1/2 (ln(17)-ln(2))

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integral from 1 to 4 of x/(1+x^2)

\int_{1}^{4} \frac{x}{1 + x ^{2}} d x

\frac{1}{2} (ln (17) - ln (2))

Decimal Notation

1.07003 \dots

Solution steps

\int_{1}^{4} \frac{x}{1 + x ^{2}} d x

Apply u-substitution

: \int_{2}^{17} \frac{1}{2 u} d u

= \int_{2}^{17} \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_{2}^{17} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{17}

Compute the boundaries:

ln (17) - ln (2)

= \frac{1}{2} (ln (17) - ln (2))

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What is the integral from 1 to 4 of x/(1+x^2) ?
The integral from 1 to 4 of x/(1+x^2) is 1/2 (ln(17)-ln(2))