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

The integral of (x+5)/(x^2(x-2)) is -7/4 ln|x|+7/4 ln|x-2|+5/(2x)+C

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

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

- \frac{7}{4} ln ∣ x ∣ + \frac{7}{4} ln ∣ x - 2 ∣ + \frac{5}{2 x} + C

Solution steps

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

Expand

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

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

\int \frac{5}{x ^{2} ( x - 2 )} d x = 5 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{1}{4} ln ∣ x - 2 ∣)

= - \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣ + 5 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{1}{4} ln ∣ x - 2 ∣)

Simplify

- \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣ + 5 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{1}{4} ln ∣ x - 2 ∣) : - \frac{7}{4} ln ∣ x ∣ + \frac{7}{4} ln ∣ x - 2 ∣ + \frac{5}{2 x}

= - \frac{7}{4} ln ∣ x ∣ + \frac{7}{4} ln ∣ x - 2 ∣ + \frac{5}{2 x}

Add a constant to the solution

= - \frac{7}{4} ln ∣ x ∣ + \frac{7}{4} ln ∣ x - 2 ∣ + \frac{5}{2 x} + C

x \to 0 lim (ln (1))

e^{- 4 x} (cos (3 x))

csc^{2} (tan (2 t))

\frac{d}{d x} (x^{m} y^{n})

(6, - 9), (- 2, - 1)

What is the integral of (x+5)/(x^2(x-2)) ?
The integral of (x+5)/(x^2(x-2)) is -7/4 ln|x|+7/4 ln|x-2|+5/(2x)+C