What is the integral from 0 to 3 of x/(1+7x^2) ?

The integral from 0 to 3 of x/(1+7x^2) is 3/7 ln(2)

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integral from 0 to 3 of x/(1+7x^2)

\int_{0}^{3} \frac{x}{1 + 7 x ^{2}} d x

\frac{3}{7} ln (2)

Decimal

0.29706 \dots

Solution steps

\int_{0}^{3} \frac{x}{1 + 7 x ^{2}} d x

Apply u-substitution

: \int_{1}^{64} \frac{1}{14 u} d u

= \int_{1}^{64} \frac{1}{14 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{14} \cdot \int_{1}^{64} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{14} [ln ∣ u ∣]_{1}^{64}

Compute the boundaries:

6 ln (2)

= \frac{1}{14} \cdot 6 ln (2)

Simplify

= \frac{3}{7} ln (2)

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What is the integral from 0 to 3 of x/(1+7x^2) ?
The integral from 0 to 3 of x/(1+7x^2) is 3/7 ln(2)