What is the integral from 1 to 7 of (15)/(x^2+5x+4) ?

The integral from 1 to 7 of (15)/(x^2+5x+4) is 5(2ln(2)+ln(5)-ln(11))

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integral from 1 to 7 of (15)/(x^2+5x+4)

\int_{1}^{7} \frac{15}{x ^{2} + 5 x + 4} d x

5 (2 ln (2) + ln (5) - ln (11))

Solution steps

\int_{1}^{7} \frac{15}{x ^{2} + 5 x + 4} d x

Take the constant out:

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

= 15 \cdot \int_{1}^{7} \frac{1}{x ^{2} + 5 x + 4} d x

Take the partial fraction of

\frac{1}{x ^{2} + 5 x + 4} : \frac{1}{3 ( x + 1 )} - \frac{1}{3 ( x + 4 )}

= 15 \cdot \int_{1}^{7} \frac{1}{3 ( x + 1 )} - \frac{1}{3 ( 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

= 15 (\int_{1}^{7} \frac{1}{3 ( x + 1 )} d x - \int_{1}^{7} \frac{1}{3 ( x + 4 )} d x)

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

\int_{1}^{7} \frac{1}{3 ( x + 4 )} d x = \frac{1}{3} (ln (11) - ln (5))

= 15 (\frac{2}{3} ln (2) - \frac{1}{3} (ln (11) - ln (5)))

Simplify

15 (\frac{2}{3} ln (2) - \frac{1}{3} (ln (11) - ln (5))) : 5 (2 ln (2) + ln (5) - ln (11))

= 5 (2 ln (2) + ln (5) - ln (11))

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What is the integral from 1 to 7 of (15)/(x^2+5x+4) ?
The integral from 1 to 7 of (15)/(x^2+5x+4) is 5(2ln(2)+ln(5)-ln(11))