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

The integral of (3x-5)/(x^2(x+5)) is 4/5 ln|x|-4/5 ln|x+5|+1/x+C

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

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

\frac{4}{5} ln ∣ x ∣ - \frac{4}{5} ln ∣ x + 5 ∣ + \frac{1}{x} + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

= \frac{4}{5} ln ∣ x ∣ - \frac{4}{5} ln ∣ x + 5 ∣ + \frac{1}{x} + C

\int (2 x - 1) (3 x + 4) d x

\int 0.002 d x

x^{2} - 6 x + 5

y^{^{''}} - 4 y^{^{'}} = 16 t, y (0) = 4, y^{^{'}} (0) = 0

1 - 2 x - 3 x^{2}

What is the integral of (3x-5)/(x^2(x+5)) ?
The integral of (3x-5)/(x^2(x+5)) is 4/5 ln|x|-4/5 ln|x+5|+1/x+C