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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (a r cot (\frac{y}{x}))

y = e^{2 e^{x}}

x \to 3 lim (x^{2} - x - 6)

\int_{1}^{8} 7 \frac{ln ( x )}{x ^{2}} d x

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