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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

f (x) = x + 15, at x = 1

(3 x^{2}) d x + (2 y - 4 x^{3} y^{- 1}) d y = 0

\int x sin^{2} (5 x) d x

\frac{d y}{d t} = 1 y

\int \frac{x + 2}{x + 1} d x

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