What is the integral of 5/(x^2+3x-4) ?

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

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integral of 5/(x^2+3x-4)

\int \frac{5}{x ^{2} + 3 x - 4} d x

ln ∣ x - 1 ∣ - ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{5}{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

= 5 \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 )}

= 5 \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

= 5 (\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 ∣

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

Simplify

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

= ln ∣ x - 1 ∣ - ln ∣ x + 4 ∣

Add a constant to the solution

= ln ∣ x - 1 ∣ - ln ∣ x + 4 ∣ + C

\int (5 x - 3)^{4} d x

y = \frac{x ^{2}}{3} + x + \frac{1}{2}

\frac{90}{x}

\int_{0}^{2} \frac{x}{( 1 + x ^{2} ) ^{2}} d x

\frac{d}{d x} ((5 - 6 x)^{5})

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