What is the integral of (361)/(x^3-19x^2) ?

The integral of (361)/(x^3-19x^2) is -ln|x|+(19)/x+ln|x-19|+C

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integral of (361)/(x^3-19x^2)

\int \frac{361}{x ^{3} - 19 x ^{2}} d x

- ln ∣ x ∣ + \frac{19}{x} + ln ∣ x - 19 ∣ + C

Solution steps

\int \frac{361}{x ^{3} - 19 x ^{2}} d x

Take the constant out:

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

= 361 \cdot \int \frac{1}{x ^{3} - 19 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 19 x ^{2}} : - \frac{1}{361 x} - \frac{1}{19 x ^{2}} + \frac{1}{361 ( x - 19 )}

= 361 \cdot \int - \frac{1}{361 x} - \frac{1}{19 x ^{2}} + \frac{1}{361 ( x - 19 )} 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

= 361 (- \int \frac{1}{361 x} d x - \int \frac{1}{19 x ^{2}} d x + \int \frac{1}{361 ( x - 19 )} d x)

\int \frac{1}{361 x} d x = \frac{1}{361} ln ∣ x ∣

\int \frac{1}{19 x ^{2}} d x = - \frac{1}{19 x}

\int \frac{1}{361 ( x - 19 )} d x = \frac{1}{361} ln ∣ x - 19 ∣

= 361 (- \frac{1}{361} ln ∣ x ∣ - (- \frac{1}{19 x}) + \frac{1}{361} ln ∣ x - 19 ∣)

Simplify

361 (- \frac{1}{361} ln ∣ x ∣ - (- \frac{1}{19 x}) + \frac{1}{361} ln ∣ x - 19 ∣) : - ln ∣ x ∣ + \frac{19}{x} + ln ∣ x - 19 ∣

= - ln ∣ x ∣ + \frac{19}{x} + ln ∣ x - 19 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{19}{x} + ln ∣ x - 19 ∣ + C

\frac{d}{d x} (- 6 sin (x))

\frac{d}{d x} (\frac{1}{20} e^{x} + x^{6} - x^{e} + e)

\frac{\partial}{\partial x} (sin^{9} (x))

\frac{\partial}{\partial y} (y \cdot cos (y - 2 x))

f (x) = x^{2} - x + 1

What is the integral of (361)/(x^3-19x^2) ?
The integral of (361)/(x^3-19x^2) is -ln|x|+(19)/x+ln|x-19|+C