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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{5}^{10} \frac{1}{2 u} d u

= 2 \cdot \int_{5}^{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

= 2 \cdot \frac{1}{2} \cdot \int_{5}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{10}

Simplify

2 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{10} : [ln ∣ u ∣]_{5}^{10}

= [ln ∣ u ∣]_{5}^{10}

Compute the boundaries:

ln (10) - ln (5)

= ln (10) - ln (5)

Simplify

= ln (2)

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\int_{- 1}^{0} - x cos (nπ x) d x

\int_{2}^{6} 2 x^{2} + 4 d x

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