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

The integral of (x^2)/((x-3)(x+2)^2) is 16/25 ln|x+2|+4/(5(x+2))+9/25 ln|x-3|+C

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

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

\frac{16}{25} ln ∣ x + 2 ∣ + \frac{4}{5 ( x + 2 )} + \frac{9}{25} ln ∣ x - 3 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{( u - 2 ) ^{2}}{u ^{2} ( u - 5 )} : \frac{16}{25 u} - \frac{4}{5 u ^{2}} + \frac{9}{25 ( u - 5 )}

= \int \frac{16}{25 u} - \frac{4}{5 u ^{2}} + \frac{9}{25 ( u - 5 )} 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{16}{25 u} d u - \int \frac{4}{5 u ^{2}} d u + \int \frac{9}{25 ( u - 5 )} d u

\int \frac{16}{25 u} d u = \frac{16}{25} ln ∣ u ∣

\int \frac{4}{5 u ^{2}} d u = - \frac{4}{5 u}

\int \frac{9}{25 ( u - 5 )} d u = \frac{9}{25} ln ∣ u - 5 ∣

= \frac{16}{25} ln ∣ u ∣ - (- \frac{4}{5 u}) + \frac{9}{25} ln ∣ u - 5 ∣

Substitute back

u = x + 2

= \frac{16}{25} ln ∣ x + 2 ∣ - (- \frac{4}{5 ( x + 2 )}) + \frac{9}{25} ln ∣ x + 2 - 5 ∣

Simplify

\frac{16}{25} ln ∣ x + 2 ∣ - (- \frac{4}{5 ( x + 2 )}) + \frac{9}{25} ln ∣ x + 2 - 5 ∣ : \frac{16}{25} ln ∣ x + 2 ∣ + \frac{4}{5 ( x + 2 )} + \frac{9}{25} ln ∣ x - 3 ∣

= \frac{16}{25} ln ∣ x + 2 ∣ + \frac{4}{5 ( x + 2 )} + \frac{9}{25} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{16}{25} ln ∣ x + 2 ∣ + \frac{4}{5 ( x + 2 )} + \frac{9}{25} ln ∣ x - 3 ∣ + C

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What is the integral of (x^2)/((x-3)(x+2)^2) ?
The integral of (x^2)/((x-3)(x+2)^2) is 16/25 ln|x+2|+4/(5(x+2))+9/25 ln|x-3|+C