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

The integral of (5x-4)/(2x^2+x-1) is -1/2 ln|2x-1|+3ln|x+1|+C

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integral of (5x-4)/(2x^2+x-1)

\int \frac{5 x - 4}{2 x ^{2} + x - 1} d x

- \frac{1}{2} ln ∣ 2 x - 1 ∣ + 3 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{5 x - 4}{2 x ^{2} + x - 1} d x

Take the partial fraction of

\frac{5 x - 4}{2 x ^{2} + x - 1} : - \frac{1}{2 x - 1} + \frac{3}{x + 1}

= \int - \frac{1}{2 x - 1} + \frac{3}{x + 1} d 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{1}{2 x - 1} d x + \int \frac{3}{x + 1} d x

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

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

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

Add a constant to the solution

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

t \to 0 lim (\frac{sin ( 4 t )}{t})

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\int x cos (\frac{x}{3}) d x

\int cos (x) csc^{6} (x) d x

What is the integral of (5x-4)/(2x^2+x-1) ?
The integral of (5x-4)/(2x^2+x-1) is -1/2 ln|2x-1|+3ln|x+1|+C