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

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

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

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

- \frac{x ^{2}}{x - 4} + 2 (x - 4 + 4 ln ∣ x - 4 ∣) - \frac{4}{x - 4} + C

Solution steps

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

Expand

\frac{x ^{2} + 4}{( x - 4 ) ^{2}} : \frac{x ^{2}}{( x - 4 ) ^{2}} + \frac{4}{( x - 4 ) ^{2}}

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

\int \frac{x ^{2}}{( x - 4 ) ^{2}} d x = - \frac{x ^{2}}{x - 4} + 2 (x - 4 + 4 ln ∣ x - 4 ∣)

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

= - \frac{x ^{2}}{x - 4} + 2 (x - 4 + 4 ln ∣ x - 4 ∣) - \frac{4}{x - 4}

Add a constant to the solution

= - \frac{x ^{2}}{x - 4} + 2 (x - 4 + 4 ln ∣ x - 4 ∣) - \frac{4}{x - 4} + C

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