What is the integral of (11x-15)/(4x^2-3x) ?

The integral of (11x-15)/(4x^2-3x) is -9/4 ln|4x-3|+5ln|x|+C

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integral of (11x-15)/(4x^2-3x)

\int \frac{11 x - 15}{4 x ^{2} - 3 x} d x

- \frac{9}{4} ln ∣ 4 x - 3 ∣ + 5 ln ∣ x ∣ + C

Solution steps

\int \frac{11 x - 15}{4 x ^{2} - 3 x} d x

Expand

\frac{11 x - 15}{4 x ^{2} - 3 x} : \frac{11 x}{4 x ^{2} - 3 x} - \frac{15}{4 x ^{2} - 3 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{11 x}{4 x ^{2} - 3 x} d x - \int \frac{15}{4 x ^{2} - 3 x} d x

\int \frac{11 x}{4 x ^{2} - 3 x} d x = \frac{11}{4} ln ∣ 4 x - 3 ∣

\int \frac{15}{4 x ^{2} - 3 x} d x = 15 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ 4 x - 3 ∣)

= \frac{11}{4} ln ∣ 4 x - 3 ∣ - 15 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ 4 x - 3 ∣)

Simplify

\frac{11}{4} ln ∣ 4 x - 3 ∣ - 15 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{3} ln ∣ 4 x - 3 ∣) : - \frac{9}{4} ln ∣ 4 x - 3 ∣ + 5 ln ∣ x ∣

= - \frac{9}{4} ln ∣ 4 x - 3 ∣ + 5 ln ∣ x ∣

Add a constant to the solution

= - \frac{9}{4} ln ∣ 4 x - 3 ∣ + 5 ln ∣ x ∣ + C

\int (sin (x))^{2} (cos (x))^{4} d x

- y^{^{'}} - y = e^{x}

n = 1 \sum \infty - 1

\frac{d y}{d x} = y + 2

f (x) = 4 sin (2 x) - 3 cos (3 x)

What is the integral of (11x-15)/(4x^2-3x) ?
The integral of (11x-15)/(4x^2-3x) is -9/4 ln|4x-3|+5ln|x|+C