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

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

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

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

\frac{1}{2} (3 ln (2) - ln (3))

Decimal Notation

0.49041 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{8} \frac{1}{2 u} d u

= \int_{3}^{8} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{3}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{3}^{8}

Compute the boundaries:

3 ln (2) - ln (3)

= \frac{1}{2} (3 ln (2) - ln (3))

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\int_{0}^{\frac{π}{2}} (x - sin (x)) d x

\int_{0}^{2} - x^{2} d x

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