What is the integral of (4x-1)/((x+9)(x+2)) ?

The integral of (4x-1)/((x+9)(x+2)) is -9/7 ln|x+2|+37/7 ln|x+9|+C

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integral of (4x-1)/((x+9)(x+2))

\int \frac{4 x - 1}{( x + 9 ) ( x + 2 )} d x

- \frac{9}{7} ln ∣ x + 2 ∣ + \frac{37}{7} ln ∣ x + 9 ∣ + C

Solution steps

\int \frac{4 x - 1}{( x + 9 ) ( x + 2 )} d x

Expand

\frac{4 x - 1}{( x + 9 ) ( x + 2 )} : \frac{4 x}{( x + 9 ) ( x + 2 )} - \frac{1}{( x + 9 ) ( x + 2 )}

Apply the Sum Rule:

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

= \int \frac{4 x}{( x + 9 ) ( x + 2 )} d x - \int \frac{1}{( x + 9 ) ( x + 2 )} d x

\int \frac{4 x}{( x + 9 ) ( x + 2 )} d x = 4 (\frac{9}{7} ln ∣ x + 9 ∣ - \frac{2}{7} ln ∣ x + 2 ∣)

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

= 4 (\frac{9}{7} ln ∣ x + 9 ∣ - \frac{2}{7} ln ∣ x + 2 ∣) - (- \frac{1}{7} ln ∣ x + 9 ∣ + \frac{1}{7} ln ∣ x + 2 ∣)

Simplify

4 (\frac{9}{7} ln ∣ x + 9 ∣ - \frac{2}{7} ln ∣ x + 2 ∣) - (- \frac{1}{7} ln ∣ x + 9 ∣ + \frac{1}{7} ln ∣ x + 2 ∣) : - \frac{9}{7} ln ∣ x + 2 ∣ + \frac{37}{7} ln ∣ x + 9 ∣

= - \frac{9}{7} ln ∣ x + 2 ∣ + \frac{37}{7} ln ∣ x + 9 ∣

Add a constant to the solution

= - \frac{9}{7} ln ∣ x + 2 ∣ + \frac{37}{7} ln ∣ x + 9 ∣ + C

\int_{3}^{4} - x^{2} + 7 x - 12 d x

\int 9 x^{2} ln (x) d x

x \to 6 + lim (ln (x^{2} - 36))

y^{^{'}} + x y = x

y = 7 e^{1 - x^{2}}, at

What is the integral of (4x-1)/((x+9)(x+2)) ?
The integral of (4x-1)/((x+9)(x+2)) is -9/7 ln|x+2|+37/7 ln|x+9|+C