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

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

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

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

2 ln (3)

Decimal Notation

2.19722 \dots

Solution steps

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

Take the constant out:

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

= 4 \cdot \int_{0}^{2} \frac{x}{2 + x ^{2}} d x

Apply u-substitution

: \int_{2}^{6} \frac{1}{2 u} d u

= 4 \cdot \int_{2}^{6} \frac{1}{2 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{2}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{6}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{6} : 2 [ln ∣ u ∣]_{2}^{6}

= 2 [ln ∣ u ∣]_{2}^{6}

Compute the boundaries:

ln (6) - ln (2)

= 2 (ln (6) - ln (2))

Simplify

= 2 ln (3)

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

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

\int_{2}^{4} (x^{2} - 2 x) d x

\int_{- 1}^{0} ln (x + 1) d x

\int_{- 1}^{1} x^{3} (3 + x^{4})^{3} d x

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