What is the integral of ((5x+6))/((x^2(x+3))) ?

The integral of ((5x+6))/((x^2(x+3))) is ln|x|-ln|x+3|-2/x+C

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integral of ((5x+6))/((x^2(x+3)))

integral

\frac{( 5 x + 6 )}{( x ^{2} ( x + 3 ) )}

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

Solution steps

\int \frac{5 x + 6}{x ^{2} ( x + 3 )} d x

Expand

\frac{5 x + 6}{x ^{2} ( x + 3 )} : \frac{5 x}{x ^{2} ( x + 3 )} + \frac{6}{x ^{2} ( x + 3 )}

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{5 x}{x ^{2} ( x + 3 )} d x + \int \frac{6}{x ^{2} ( x + 3 )} d x

\int \frac{5 x}{x ^{2} ( x + 3 )} d x = 5 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣)

\int \frac{6}{x ^{2} ( x + 3 )} d x = 6 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

= 5 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣) + 6 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

Simplify

5 (\frac{1}{3} ln ∣ x ∣ - \frac{1}{3} ln ∣ x + 3 ∣) + 6 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣) : ln ∣ x ∣ - ln ∣ x + 3 ∣ - \frac{2}{x}

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

Add a constant to the solution

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

x \cdot (x + 5)^{8}

tan^{3} (2 x + 1)

(sin (x) + 3)^{- 1}

e^{- 5 x^{5}}

\frac{- 1}{x + 1 ^{2}}

What is the integral of ((5x+6))/((x^2(x+3))) ?
The integral of ((5x+6))/((x^2(x+3))) is ln|x|-ln|x+3|-2/x+C