What is the integral of (x^2)/((x-4)(x-5)^2) ?

The integral of (x^2)/((x-4)(x-5)^2) is 16ln|x-4|-15ln|x-5|-(25)/(x-5)+C

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integral of (x^2)/((x-4)(x-5)^2)

\int \frac{x ^{2}}{( x - 4 ) ( x - 5 ) ^{2}} d x

16 ln ∣ x - 4 ∣ - 15 ln ∣ x - 5 ∣ - \frac{25}{x - 5} + C

Solution steps

\int \frac{x ^{2}}{( x - 4 ) ( x - 5 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 4 ) ( x - 5 ) ^{2}} : \frac{16}{x - 4} - \frac{15}{x - 5} + \frac{25}{( x - 5 ) ^{2}}

= \int \frac{16}{x - 4} - \frac{15}{x - 5} + \frac{25}{( x - 5 ) ^{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{16}{x - 4} d x - \int \frac{15}{x - 5} d x + \int \frac{25}{( x - 5 ) ^{2}} d x

\int \frac{16}{x - 4} d x = 16 ln ∣ x - 4 ∣

\int \frac{15}{x - 5} d x = 15 ln ∣ x - 5 ∣

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

= 16 ln ∣ x - 4 ∣ - 15 ln ∣ x - 5 ∣ - \frac{25}{x - 5}

Add a constant to the solution

= 16 ln ∣ x - 4 ∣ - 15 ln ∣ x - 5 ∣ - \frac{25}{x - 5} + C

What is the integral of (x^2)/((x-4)(x-5)^2) ?
The integral of (x^2)/((x-4)(x-5)^2) is 16ln|x-4|-15ln|x-5|-(25)/(x-5)+C