What is the integral of (18)/(x^2+4x-5) ?

The integral of (18)/(x^2+4x-5) is 18(1/6 ln|x-1|-1/6 ln|x+5|)+C

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integral of (18)/(x^2+4x-5)

\int \frac{18}{x ^{2} + 4 x - 5} d x

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

Solution steps

\int \frac{18}{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

= 18 \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 )}

= 18 \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

= 18 (\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 ∣

= 18 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣)

Add a constant to the solution

= 18 (\frac{1}{6} ln ∣ x - 1 ∣ - \frac{1}{6} ln ∣ x + 5 ∣) + C

in v erse l a pl a ce \frac{1}{( s ^{3} + 1 ) ( s ^{2} + 1 )}

d er i v a t i v e x^{2} ln (4 x + 3)

\int x sin (π x) d x

\frac{\partial}{\partial x} (k ln (x^{2} + y^{2}))

n = 1 \sum \infty (- \frac{1}{2})^{n + 1}

What is the integral of (18)/(x^2+4x-5) ?
The integral of (18)/(x^2+4x-5) is 18(1/6 ln|x-1|-1/6 ln|x+5|)+C