What is the integral of (x+1)/(4x^2-x) ?

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

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integral of (x+1)/(4x^2-x)

\int \frac{x + 1}{4 x ^{2} - x} d x

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

Solution steps

\int \frac{x + 1}{4 x ^{2} - x} d x

Expand

\frac{x + 1}{4 x ^{2} - x} : \frac{x}{4 x ^{2} - x} + \frac{1}{4 x ^{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{x}{4 x ^{2} - x} d x + \int \frac{1}{4 x ^{2} - x} d x

\int \frac{x}{4 x ^{2} - x} d x = \frac{1}{4} ln ∣ 4 x - 1 ∣

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

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

Add similar elements:

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

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

Add a constant to the solution

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

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