What is the integral of (x^2)/((x-6)(x^2-36)) ?

The integral of (x^2)/((x-6)(x^2-36)) is 3/4 ln|x-6|-3/(x-6)+1/4 ln|x+6|+C

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integral of (x^2)/((x-6)(x^2-36))

\int \frac{x ^{2}}{( x - 6 ) ( x ^{2} - 36 )} d x

\frac{3}{4} ln ∣ x - 6 ∣ - \frac{3}{x - 6} + \frac{1}{4} ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{x ^{2}}{( x - 6 ) ( x ^{2} - 36 )} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 6 ) ( x ^{2} - 36 )} : \frac{3}{4 ( x - 6 )} + \frac{3}{( x - 6 ) ^{2}} + \frac{1}{4 ( x + 6 )}

= \int \frac{3}{4 ( x - 6 )} + \frac{3}{( x - 6 ) ^{2}} + \frac{1}{4 ( x + 6 )} 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{3}{4 ( x - 6 )} d x + \int \frac{3}{( x - 6 ) ^{2}} d x + \int \frac{1}{4 ( x + 6 )} d x

\int \frac{3}{4 ( x - 6 )} d x = \frac{3}{4} ln ∣ x - 6 ∣

\int \frac{3}{( x - 6 ) ^{2}} d x = - \frac{3}{x - 6}

\int \frac{1}{4 ( x + 6 )} d x = \frac{1}{4} ln ∣ x + 6 ∣

= \frac{3}{4} ln ∣ x - 6 ∣ - \frac{3}{x - 6} + \frac{1}{4} ln ∣ x + 6 ∣

Add a constant to the solution

= \frac{3}{4} ln ∣ x - 6 ∣ - \frac{3}{x - 6} + \frac{1}{4} ln ∣ x + 6 ∣ + C

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What is the integral of (x^2)/((x-6)(x^2-36)) ?
The integral of (x^2)/((x-6)(x^2-36)) is 3/4 ln|x-6|-3/(x-6)+1/4 ln|x+6|+C