What is the integral of 2/(x^2+3x-4) ?

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

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integral of 2/(x^2+3x-4)

\int \frac{2}{x ^{2} + 3 x - 4} d x

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

Solution steps

\int \frac{2}{x ^{2} + 3 x - 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{2} + 3 x - 4} d x

Take the partial fraction of

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

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

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

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

\int \frac{1}{5 ( x + 4 )} d x = \frac{1}{5} ln ∣ x + 4 ∣

= 2 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣)

Add a constant to the solution

= 2 (\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣) + C

\int_{0}^{π} cos (\frac{x}{2}) d x

x \to 1 - lim (\frac{∣ - 1 ∣}{- 1})

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 25})

\int cos (10 x) d x

\frac{\partial}{\partial x} (x e^{2 x + 3 y})

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