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

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

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

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

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

Decimal Notation

0.29389 \dots

Solution steps

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

Complete the square

x^{2} + 2 x : (x + 1)^{2} - 1

= \int_{1}^{3} \frac{1}{( x + 1 ) ^{2} - 1} d x

Apply u-substitution

: \int_{2}^{4} \frac{1}{u ^{2} - 1} d u

= \int_{2}^{4} \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \int_{2}^{4} \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= - \int_{2}^{4} \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{2}^{4}

Simplify

- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{2}^{4} : - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{2}^{4}

= - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{2}^{4}

Compute the boundaries:

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

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

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