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

The integral from 1 to 2 of (x^2)/(6x^3+1) is 1/18 ln(7)

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

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

\frac{1}{18} ln (7)

Decimal Notation

0.10810 \dots

Solution steps

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

Apply u-substitution

: \int_{7}^{49} \frac{1}{18 u} d u

= \int_{7}^{49} \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_{7}^{49} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{18} [ln ∣ u ∣]_{7}^{49}

Compute the boundaries:

ln (7)

= \frac{1}{18} ln (7)

\int_{0}^{\frac{π}{2}} (x + 1) cos (x) d x

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

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\int_{1}^{l n (2)} x \cdot e^{4 x} d x

\int_{0}^{r} (b r + c) 2 π r d r

What is the integral from 1 to 2 of (x^2)/(6x^3+1) ?
The integral from 1 to 2 of (x^2)/(6x^3+1) is 1/18 ln(7)