What is the integral from 0 to 1 of (x^2)/(4x^3+1) ?

The integral from 0 to 1 of (x^2)/(4x^3+1) is 1/12 ln(5)

Popular >

integral from 0 to 1 of (x^2)/(4x^3+1)

\int_{0}^{1} \frac{x ^{2}}{4 x ^{3} + 1} d x

\frac{1}{12} ln (5)

Decimal

0.13411 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{4 x ^{3} + 1} d x

Apply u-substitution

: \int_{1}^{5} \frac{1}{12 u} d u

= \int_{1}^{5} \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}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{12} [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= \frac{1}{12} ln (5)

\int_{0}^{1} sin (a x) d x

\int_{6}^{7} 1

\int_{- 1}^{1} x^{3} - \frac{3}{5} x d x

\int_{1}^{2} (x + 1)^{2} d x

\int_{- 2}^{0} x^{3} 16 - x^{4} d x

What is the integral from 0 to 1 of (x^2)/(4x^3+1) ?
The integral from 0 to 1 of (x^2)/(4x^3+1) is 1/12 ln(5)