What is the integral of (7x+6)/(x^2(x+2)) ?

The integral of (7x+6)/(x^2(x+2)) is 2ln|x|-2ln|x+2|-3/x+C

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integral of (7x+6)/(x^2(x+2))

\int \frac{7 x + 6}{x ^{2} ( x + 2 )} d x

2 ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ - \frac{3}{x} + C

Solution steps

\int \frac{7 x + 6}{x ^{2} ( x + 2 )} d x

Expand

\frac{7 x + 6}{x ^{2} ( x + 2 )} : \frac{7 x}{x ^{2} ( x + 2 )} + \frac{6}{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{7 x}{x ^{2} ( x + 2 )} d x + \int \frac{6}{x ^{2} ( x + 2 )} d x

\int \frac{7 x}{x ^{2} ( x + 2 )} d x = 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣)

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

= 7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + 6 (- \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{1}{4} ln ∣ x + 2 ∣)

Simplify

7 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣) + 6 (- \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{1}{4} ln ∣ x + 2 ∣) : 2 ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ - \frac{3}{x}

= 2 ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ - \frac{3}{x}

Add a constant to the solution

= 2 ln ∣ x ∣ - 2 ln ∣ x + 2 ∣ - \frac{3}{x} + C

y = 8 x^{2} + 8, (2, 40)

\frac{\partial}{\partial z} (x e^{y - z^{2}})

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

x \to 5 + lim (5 x - 5)

\frac{\partial}{\partial t} (t^{2} e^{- x})

What is the integral of (7x+6)/(x^2(x+2)) ?
The integral of (7x+6)/(x^2(x+2)) is 2ln|x|-2ln|x+2|-3/x+C