What is the integral from 1 to 3 of 1/(x^2+4x) ?

The integral from 1 to 3 of 1/(x^2+4x) is (ln(3)-ln(7)+ln(5))/4

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integral from 1 to 3 of 1/(x^2+4x)

\int_{1}^{3} \frac{1}{x ^{2} + 4 x} d x

\frac{ln ( 3 ) - ln ( 7 ) + ln ( 5 )}{4}

Decimal Notation

0.19053 \dots

Solution steps

\int_{1}^{3} \frac{1}{x ^{2} + 4 x} d x

Take the partial fraction of

\frac{1}{x ^{2} + 4 x} : \frac{1}{4 x} - \frac{1}{4 ( x + 4 )}

= \int_{1}^{3} \frac{1}{4 x} - \frac{1}{4 ( x + 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} \frac{1}{4 x} d x - \int_{1}^{3} \frac{1}{4 ( x + 4 )} d x

\int_{1}^{3} \frac{1}{4 x} d x = \frac{1}{4} ln (3)

\int_{1}^{3} \frac{1}{4 ( x + 4 )} d x = \frac{1}{4} (ln (7) - ln (5))

= \frac{1}{4} ln (3) - \frac{1}{4} (ln (7) - ln (5))

Simplify

\frac{1}{4} ln (3) - \frac{1}{4} (ln (7) - ln (5)) : \frac{ln ( 3 ) - ln ( 7 ) + ln ( 5 )}{4}

= \frac{ln ( 3 ) - ln ( 7 ) + ln ( 5 )}{4}

\int_{1}^{7} (- \frac{1}{3} x^{2} + 3 x) d x

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What is the integral from 1 to 3 of 1/(x^2+4x) ?
The integral from 1 to 3 of 1/(x^2+4x) is (ln(3)-ln(7)+ln(5))/4