What is the integral of 2/(x^2+11x+18) ?

The integral of 2/(x^2+11x+18) is 2(1/7 ln|x+2|-1/7 ln|x+9|)+C

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integral of 2/(x^2+11x+18)

\int \frac{2}{x ^{2} + 11 x + 18} d x

2 (\frac{1}{7} ln ∣ x + 2 ∣ - \frac{1}{7} ln ∣ x + 9 ∣) + C

Solution steps

\int \frac{2}{x ^{2} + 11 x + 18} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{2} + 11 x + 18} d x

Take the partial fraction of

\frac{1}{x ^{2} + 11 x + 18} : \frac{1}{7 ( x + 2 )} - \frac{1}{7 ( x + 9 )}

= 2 \cdot \int \frac{1}{7 ( x + 2 )} - \frac{1}{7 ( x + 9 )} 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

= 2 (\int \frac{1}{7 ( x + 2 )} d x - \int \frac{1}{7 ( x + 9 )} d x)

\int \frac{1}{7 ( x + 2 )} d x = \frac{1}{7} ln ∣ x + 2 ∣

\int \frac{1}{7 ( x + 9 )} d x = \frac{1}{7} ln ∣ x + 9 ∣

= 2 (\frac{1}{7} ln ∣ x + 2 ∣ - \frac{1}{7} ln ∣ x + 9 ∣)

Add a constant to the solution

= 2 (\frac{1}{7} ln ∣ x + 2 ∣ - \frac{1}{7} ln ∣ x + 9 ∣) + C

\int 5 cos (3 x) - 4 sin (5 x) d x

lo g_{5} (x)

\frac{1}{3} x^{\frac{1}{2}} - x^{\frac{3}{2}}, x = 0, x = \frac{1}{3}

\frac{d ^{2}}{d x ^{2}} (x 9 - x)

x \to 4 - lim (\frac{7}{x - 4})

What is the integral of 2/(x^2+11x+18) ?
The integral of 2/(x^2+11x+18) is 2(1/7 ln|x+2|-1/7 ln|x+9|)+C