What is the integral of ((3x-5))/(x^2+4x) ?

The integral of ((3x-5))/(x^2+4x) is -5/4 ln|x|+17/4 ln|x+4|+C

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integral of ((3x-5))/(x^2+4x)

\int \frac{( 3 x - 5 )}{x ^{2} + 4 x} d x

- \frac{5}{4} ln ∣ x ∣ + \frac{17}{4} ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{3 x - 5}{x ^{2} + 4 x} d x

Expand

\frac{3 x - 5}{x ^{2} + 4 x} : \frac{3 x}{x ^{2} + 4 x} - \frac{5}{x ^{2} + 4 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{3 x}{x ^{2} + 4 x} d x - \int \frac{5}{x ^{2} + 4 x} d x

\int \frac{3 x}{x ^{2} + 4 x} d x = 3 ln ∣ x + 4 ∣

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

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

Simplify

3 ln ∣ x + 4 ∣ - 5 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣) : - \frac{5}{4} ln ∣ x ∣ + \frac{17}{4} ln ∣ x + 4 ∣

= - \frac{5}{4} ln ∣ x ∣ + \frac{17}{4} ln ∣ x + 4 ∣

Add a constant to the solution

= - \frac{5}{4} ln ∣ x ∣ + \frac{17}{4} ln ∣ x + 4 ∣ + C

f (x) = 3 x^{2} - 7

\int e^{x} sinh (x) d x

(- cos (x))^{^{'}}

\frac{\partial}{\partial x} (arccos (x))

\int_{3}^{\infty} \frac{1}{x ^{2} - 4} d x

What is the integral of ((3x-5))/(x^2+4x) ?
The integral of ((3x-5))/(x^2+4x) is -5/4 ln|x|+17/4 ln|x+4|+C