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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

Add a constant to the solution

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

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