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

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

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

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

2 ln (2)

Decimal

1.38629 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 6 \cdot \int_{1}^{4} \frac{1}{6 u} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{1}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{4}

Simplify

6 \cdot \frac{1}{6} [ln ∣ u ∣]_{1}^{4} : [ln ∣ u ∣]_{1}^{4}

= [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= 2 ln (2)

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