What is the integral of (272x^2+4)/(x^2(x-4)) ?

The integral of (272x^2+4)/(x^2(x-4)) is -1/4 ln|x|+1089/4 ln|x-4|+1/x+C

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integral of (272x^2+4)/(x^2(x-4))

\int \frac{272 x ^{2} + 4}{x ^{2} ( x - 4 )} d x

- \frac{1}{4} ln ∣ x ∣ + \frac{1089}{4} ln ∣ x - 4 ∣ + \frac{1}{x} + C

Solution steps

\int \frac{272 x ^{2} + 4}{x ^{2} ( x - 4 )} d x

Expand

\frac{272 x ^{2} + 4}{x ^{2} ( x - 4 )} : \frac{272 x ^{2}}{x ^{2} ( x - 4 )} + \frac{4}{x ^{2} ( x - 4 )}

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

\int \frac{272 x ^{2}}{x ^{2} ( x - 4 )} d x = 272 ln ∣ x - 4 ∣

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

= 272 ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{4} ln ∣ x - 4 ∣

Simplify

272 ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{x} + \frac{1}{4} ln ∣ x - 4 ∣ : - \frac{1}{4} ln ∣ x ∣ + \frac{1089}{4} ln ∣ x - 4 ∣ + \frac{1}{x}

= - \frac{1}{4} ln ∣ x ∣ + \frac{1089}{4} ln ∣ x - 4 ∣ + \frac{1}{x}

Add a constant to the solution

= - \frac{1}{4} ln ∣ x ∣ + \frac{1089}{4} ln ∣ x - 4 ∣ + \frac{1}{x} + C

(2 x y - 3 x^{2}) d x + (x^{2} - 2 y) d y = 0

\frac{d y}{d x} + y^{5} x + 3 y = 0

y^{2} \frac{d y}{d x} + y^{3} x^{- 1} = \frac{cos ( x )}{x}

\frac{2}{( s - 1 ) ^{2}}

\int \frac{9 x}{3 x + 1} d x

What is the integral of (272x^2+4)/(x^2(x-4)) ?
The integral of (272x^2+4)/(x^2(x-4)) is -1/4 ln|x|+1089/4 ln|x-4|+1/x+C