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

The integral of (2x-3)/(x^2-3x+5x) is 7/2 ln|x+2|-3/2 ln|x|+C

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

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

\frac{7}{2} ln ∣ x + 2 ∣ - \frac{3}{2} ln ∣ x ∣ + C

Solution steps

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

Simplify

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

Expand

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

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

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

= 2 ln ∣ x + 2 ∣ - (- 3 (\frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{2} ln ∣ x ∣))

Simplify

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

= \frac{7}{2} ln ∣ x + 2 ∣ - \frac{3}{2} ln ∣ x ∣

Add a constant to the solution

= \frac{7}{2} ln ∣ x + 2 ∣ - \frac{3}{2} ln ∣ x ∣ + C

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What is the integral of (2x-3)/(x^2-3x+5x) ?
The integral of (2x-3)/(x^2-3x+5x) is 7/2 ln|x+2|-3/2 ln|x|+C