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

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

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integral from 1 to 2 of (5x)/(x^2+1)

\int_{1}^{2} \frac{5 x}{x ^{2} + 1} d x

\frac{5}{2} (ln (5) - ln (2))

Decimal Notation

2.29072 \dots

Solution steps

\int_{1}^{2} \frac{5 x}{x ^{2} + 1} d x

Take the constant out:

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

= 5 \cdot \int_{1}^{2} \frac{x}{x ^{2} + 1} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

= \frac{5}{2} [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

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

\int \frac{( x + 2 )}{4 x - x ^{2}} d x

\frac{1}{60} w h

\int (x + 1) (x^{2} + 2 x)^{2} d x

\frac{d}{d x} (\frac{64}{x})

f (x) = 8 x^{2} + 5 x, at x = 3

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