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

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

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

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

3 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣) + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} + 4 x - 5} d x

Take the partial fraction of

\frac{1}{x ^{2} + 4 x - 5} : \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( x + 5 )}

= 3 \cdot \int \frac{1}{6 ( x - 1 )} - \frac{1}{6 ( x + 5 )} 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

= 3 (\int \frac{1}{6 ( x - 1 )} d x - \int \frac{1}{6 ( x + 5 )} d x)

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

\int \frac{1}{6 ( x + 5 )} d x = \frac{1}{6} ln ∣ x + 5 ∣

= 3 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣)

Add a constant to the solution

= 3 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣) + C

x \to 0 + lim ((1 + x)^{\frac{2}{x}})

f (x) = e^{x^{5} + 2 x}

y = x^{2} y^{3} + x^{3} y^{2}

\frac{d y}{d x} = e^{(6 x - 3 y)}

\frac{d}{d x} (\frac{1}{( 16 - x ^{2} )})

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