What is the integral of (3x+1)/((x-2)(x+3)^2) ?

The integral of (3x+1)/((x-2)(x+3)^2) is -7/25 ln|x+3|+7/25 ln|x-2|-8/(5(x+3))+C

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integral of (3x+1)/((x-2)(x+3)^2)

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

- \frac{7}{25} ln ∣ x + 3 ∣ + \frac{7}{25} ln ∣ x - 2 ∣ - \frac{8}{5 ( x + 3 )} + C

Solution steps

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

Expand

\frac{3 x + 1}{( x - 2 ) ( x + 3 ) ^{2}} : \frac{3 x}{( x - 2 ) ( x + 3 ) ^{2}} + \frac{1}{( x - 2 ) ( x + 3 ) ^{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{3 x}{( x - 2 ) ( x + 3 ) ^{2}} d x + \int \frac{1}{( x - 2 ) ( x + 3 ) ^{2}} d x

\int \frac{3 x}{( x - 2 ) ( x + 3 ) ^{2}} d x = 3 (- \frac{2}{25} ln ∣ x + 3 ∣ - \frac{3}{5 ( x + 3 )} + \frac{2}{25} ln ∣ x - 2 ∣)

\int \frac{1}{( x - 2 ) ( x + 3 ) ^{2}} d x = \frac{1}{25} ln ∣ x - 2 ∣ - \frac{1}{25} ln ∣ x + 3 ∣ + \frac{1}{5 ( x + 3 )}

= 3 (- \frac{2}{25} ln ∣ x + 3 ∣ - \frac{3}{5 ( x + 3 )} + \frac{2}{25} ln ∣ x - 2 ∣) + \frac{1}{25} ln ∣ x - 2 ∣ - \frac{1}{25} ln ∣ x + 3 ∣ + \frac{1}{5 ( x + 3 )}

Simplify

3 (- \frac{2}{25} ln ∣ x + 3 ∣ - \frac{3}{5 ( x + 3 )} + \frac{2}{25} ln ∣ x - 2 ∣) + \frac{1}{25} ln ∣ x - 2 ∣ - \frac{1}{25} ln ∣ x + 3 ∣ + \frac{1}{5 ( x + 3 )} : - \frac{7}{25} ln ∣ x + 3 ∣ + \frac{7}{25} ln ∣ x - 2 ∣ - \frac{8}{5 ( x + 3 )}

= - \frac{7}{25} ln ∣ x + 3 ∣ + \frac{7}{25} ln ∣ x - 2 ∣ - \frac{8}{5 ( x + 3 )}

Add a constant to the solution

= - \frac{7}{25} ln ∣ x + 3 ∣ + \frac{7}{25} ln ∣ x - 2 ∣ - \frac{8}{5 ( x + 3 )} + C

\frac{\partial}{\partial x} (\frac{( y + z )}{x})

\frac{d}{d x} (k (sec (x) + tan (x)))

(4 + x) y^{^{'}} = 3 y

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

\frac{d y}{d x} = 1 + y

What is the integral of (3x+1)/((x-2)(x+3)^2) ?
The integral of (3x+1)/((x-2)(x+3)^2) is -7/25 ln|x+3|+7/25 ln|x-2|-8/(5(x+3))+C