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

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

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

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

\frac{3}{2} ln (2)

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \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

= 3 \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 ∣)

= 3 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{10}

Simplify

3 \cdot \frac{1}{2} [ln ∣ u ∣]_{5}^{10} : \frac{3}{2} [ln ∣ u ∣]_{5}^{10}

= \frac{3}{2} [ln ∣ u ∣]_{5}^{10}

Compute the boundaries:

ln (10) - ln (5)

= \frac{3}{2} (ln (10) - ln (5))

Simplify

= \frac{3}{2} ln (2)

\int_{12}^{37} x x^{2} - 144 d x

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

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

\int_{- 1}^{5} cos (x) d x

\int_{- 2}^{- 1} 3 e^{- 5 x + 8} d x

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