What is the integral of (3-x)/(2x^2+x-1) ?

The integral of (3-x)/(2x^2+x-1) is 5/6 ln|2x-1|-4/3 ln|x+1|+C

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integral of (3-x)/(2x^2+x-1)

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

\frac{5}{6} ln ∣ 2 x - 1 ∣ - \frac{4}{3} ln ∣ x + 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int \frac{5}{3 ( 2 x - 1 )} - \frac{4}{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{5}{3 ( 2 x - 1 )} d x - \int \frac{4}{3 ( x + 1 )} d x

\int \frac{5}{3 ( 2 x - 1 )} d x = \frac{5}{6} ln ∣ 2 x - 1 ∣

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

= \frac{5}{6} ln ∣ 2 x - 1 ∣ - \frac{4}{3} ln ∣ x + 1 ∣

Add a constant to the solution

= \frac{5}{6} ln ∣ 2 x - 1 ∣ - \frac{4}{3} ln ∣ x + 1 ∣ + C

x v^{'} + v = \frac{2 v}{1 - v ^{2}}

f (x) = 2 x + 77, at x = 2

x \to 0 - lim (\frac{( 6 x )}{x ^{4}})

y^{^{''}} - 8 y^{^{'}} + 7 y = 0

\int_{0}^{5} (140 x^{2} + 9800) d x

What is the integral of (3-x)/(2x^2+x-1) ?
The integral of (3-x)/(2x^2+x-1) is 5/6 ln|2x-1|-4/3 ln|x+1|+C