What is the integral from 0 to 3 of x/(1+6x^2) ?

The integral from 0 to 3 of x/(1+6x^2) is 1/12 ln(55)

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integral from 0 to 3 of x/(1+6x^2)

\int_{0}^{3} \frac{x}{1 + 6 x ^{2}} d x

\frac{1}{12} ln (55)

Decimal

0.33394 \dots

Solution steps

\int_{0}^{3} \frac{x}{1 + 6 x ^{2}} d x

Apply u-substitution

: \int_{1}^{55} \frac{1}{12 u} d u

= \int_{1}^{55} \frac{1}{12 u} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{1}^{55} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{12} [ln ∣ u ∣]_{1}^{55}

Compute the boundaries:

ln (55)

= \frac{1}{12} ln (55)

\int_{0.5}^{1.5} - 2 x + 4 d x

\int_{0}^{y} 6 y d y

\int_{2}^{3} 7 x d x

\int_{- 2}^{0} x^{3} - 3 x^{2} - 10 x d x

\int_{0}^{x^{3}} f (t) d t

What is the integral from 0 to 3 of x/(1+6x^2) ?
The integral from 0 to 3 of x/(1+6x^2) is 1/12 ln(55)