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

The integral from 0 to 1 of (x^3)/(3x^4+1) is 1/6 ln(2)

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

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

\frac{1}{6} ln (2)

Decimal Notation

0.11552 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{4} \frac{1}{12 u} d u

= \int_{1}^{4} \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}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{12} [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= \frac{1}{12} \cdot 2 ln (2)

Simplify

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

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

\int_{0}^{1} 2 x e^{3 x} d x

\int_{- 2}^{1} x^{2} - x + 2 d x

\int_{0}^{2} x^{2} - 5 d x

\int_{- 1}^{1} 2 y^{2} d y

What is the integral from 0 to 1 of (x^3)/(3x^4+1) ?
The integral from 0 to 1 of (x^3)/(3x^4+1) is 1/6 ln(2)