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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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

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

x \to 0 lim (\frac{sin ( 4 x )}{( 2 x )})

\int_{2}^{\infty} y e^{- 5 y} d y

(\frac{1}{14 x ^{2}}), x, (\frac{x}{4})

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