What is the integral from 0 to 3 of x/(1+9x^2) ?

The integral from 0 to 3 of x/(1+9x^2) is 1/18 ln(82)

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integral from 0 to 3 of x/(1+9x^2)

\int_{0}^{3} \frac{x}{1 + 9 x ^{2}} d x

\frac{1}{18} ln (82)

Decimal

0.24481 \dots

Solution steps

\int_{0}^{3} \frac{x}{1 + 9 x ^{2}} d x

Apply u-substitution

: \int_{1}^{82} \frac{1}{18 u} d u

= \int_{1}^{82} \frac{1}{18 u} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int_{1}^{82} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{18} [ln ∣ u ∣]_{1}^{82}

Compute the boundaries:

ln (82)

= \frac{1}{18} ln (82)

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What is the integral from 0 to 3 of x/(1+9x^2) ?
The integral from 0 to 3 of x/(1+9x^2) is 1/18 ln(82)