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

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

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

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

2 ln (5)

Decimal

3.21887 \dots

Solution steps

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

Apply u-substitution

= 4 \cdot \int_{2}^{10} \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}^{10} \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}^{10}

Simplify

4 \cdot \frac{1}{2} [ln ∣ u ∣]_{2}^{10} : 2 [ln ∣ u ∣]_{2}^{10}

= 2 [ln ∣ u ∣]_{2}^{10}

Compute the boundaries:

ln (10) - ln (2)

= 2 (ln (10) - ln (2))

Simplify

= 2 ln (5)

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