What is the integral of (10-2x)/(x^2+5x) ?

The integral of (10-2x)/(x^2+5x) is 2ln|x|-4ln|x+5|+C

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integral of (10-2x)/(x^2+5x)

\int \frac{10 - 2 x}{x ^{2} + 5 x} d x

2 ln ∣ x ∣ - 4 ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{10 - 2 x}{x ^{2} + 5 x} d x

Expand

\frac{10 - 2 x}{x ^{2} + 5 x} : \frac{10}{x ^{2} + 5 x} - \frac{2 x}{x ^{2} + 5 x}

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

\int \frac{10}{x ^{2} + 5 x} d x = 10 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{5} ln ∣ x + 5 ∣)

\int \frac{2 x}{x ^{2} + 5 x} d x = 2 ln ∣ x + 5 ∣

= 10 (\frac{1}{5} ln ∣ x ∣ - \frac{1}{5} ln ∣ x + 5 ∣) - 2 ln ∣ x + 5 ∣

Simplify

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

= 2 ln ∣ x ∣ - 4 ln ∣ x + 5 ∣

Add a constant to the solution

= 2 ln ∣ x ∣ - 4 ln ∣ x + 5 ∣ + C

\int 1.05 (x + 180)^{0.05} d x

\int \frac{cos ( x )}{( cos ( x ) + sin ( x ) )} d x

\int x^{2} cot (x^{3}) csc (x^{3}) d x

\int - 5 e^{5 t} + 4 d t

ln (x) + 1

What is the integral of (10-2x)/(x^2+5x) ?
The integral of (10-2x)/(x^2+5x) is 2ln|x|-4ln|x+5|+C