What is the integral from 0 to 4 of (12)/(x^2+5x+6) ?

The integral from 0 to 4 of (12)/(x^2+5x+6) is 12(2ln(3)-ln(7))

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integral from 0 to 4 of (12)/(x^2+5x+6)

\int_{0}^{4} \frac{12}{x ^{2} + 5 x + 6} d x

12 (2 ln (3) - ln (7))

Decimal

3.01577 \dots

Solution steps

\int_{0}^{4} \frac{12}{x ^{2} + 5 x + 6} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{4} \frac{1}{x ^{2} + 5 x + 6} d x

Take the partial fraction of

\frac{1}{x ^{2} + 5 x + 6} : \frac{1}{x + 2} - \frac{1}{x + 3}

= 12 \cdot \int_{0}^{4} \frac{1}{x + 2} - \frac{1}{x + 3} 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

= 12 (\int_{0}^{4} \frac{1}{x + 2} d x - \int_{0}^{4} \frac{1}{x + 3} d x)

\int_{0}^{4} \frac{1}{x + 2} d x = ln (3)

\int_{0}^{4} \frac{1}{x + 3} d x = ln (7) - ln (3)

= 12 (ln (3) - (ln (7) - ln (3)))

Simplify

= 12 (2 ln (3) - ln (7))

\int \frac{4}{x ^{2} + 8 x} d x

\frac{d}{d x} (x e^{x y} cos (2 x) - 3)

\int_{- 1}^{1} \frac{3 x}{( x + 6 ) ^{2}} d x

\int 4 x y d x

e^{x}

What is the integral from 0 to 4 of (12)/(x^2+5x+6) ?
The integral from 0 to 4 of (12)/(x^2+5x+6) is 12(2ln(3)-ln(7))