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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u - 5 ) ^{2}} d u

= \int \frac{1}{u ( u - 5 ) ^{2}} d u

Take the partial fraction of

\frac{1}{u ( u - 5 ) ^{2}} : \frac{1}{25 u} - \frac{1}{25 ( u - 5 )} + \frac{1}{5 ( u - 5 ) ^{2}}

= \int \frac{1}{25 u} - \frac{1}{25 ( u - 5 )} + \frac{1}{5 ( u - 5 ) ^{2}} d u

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{1}{25 u} d u - \int \frac{1}{25 ( u - 5 )} d u + \int \frac{1}{5 ( u - 5 ) ^{2}} d u

\int \frac{1}{25 u} d u = \frac{1}{25} ln ∣ u ∣

\int \frac{1}{25 ( u - 5 )} d u = \frac{1}{25} ln ∣ u - 5 ∣

\int \frac{1}{5 ( u - 5 ) ^{2}} d u = - \frac{1}{5 ( u - 5 )}

= \frac{1}{25} ln ∣ u ∣ - \frac{1}{25} ln ∣ u - 5 ∣ - \frac{1}{5 ( u - 5 )}

Substitute back

u = x + 3

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

Simplify

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

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

Add a constant to the solution

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

\int (3 e^{x} + 2 sin (x)) d x

f (x) = 3 x^{2} + 2 x - 4

y = (1 + 5 x)^{9}, (0, 1)

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

x \to 8 lim (\frac{x}{x - 8})

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