What is the integral of (81)/(x^3-9x^2) ?

The integral of (81)/(x^3-9x^2) is -ln|x|+9/x+ln|x-9|+C

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integral of (81)/(x^3-9x^2)

\int \frac{81}{x ^{3} - 9 x ^{2}} d x

- ln ∣ x ∣ + \frac{9}{x} + ln ∣ x - 9 ∣ + C

Solution steps

\int \frac{81}{x ^{3} - 9 x ^{2}} d x

Take the constant out:

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

= 81 \cdot \int \frac{1}{x ^{3} - 9 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 9 x ^{2}} : - \frac{1}{81 x} - \frac{1}{9 x ^{2}} + \frac{1}{81 ( x - 9 )}

= 81 \cdot \int - \frac{1}{81 x} - \frac{1}{9 x ^{2}} + \frac{1}{81 ( x - 9 )} 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

= 81 (- \int \frac{1}{81 x} d x - \int \frac{1}{9 x ^{2}} d x + \int \frac{1}{81 ( x - 9 )} d x)

\int \frac{1}{81 x} d x = \frac{1}{81} ln ∣ x ∣

\int \frac{1}{9 x ^{2}} d x = - \frac{1}{9 x}

\int \frac{1}{81 ( x - 9 )} d x = \frac{1}{81} ln ∣ x - 9 ∣

= 81 (- \frac{1}{81} ln ∣ x ∣ - (- \frac{1}{9 x}) + \frac{1}{81} ln ∣ x - 9 ∣)

Simplify

81 (- \frac{1}{81} ln ∣ x ∣ - (- \frac{1}{9 x}) + \frac{1}{81} ln ∣ x - 9 ∣) : - ln ∣ x ∣ + \frac{9}{x} + ln ∣ x - 9 ∣

= - ln ∣ x ∣ + \frac{9}{x} + ln ∣ x - 9 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{9}{x} + ln ∣ x - 9 ∣ + C

\frac{x ^{2} + 1}{x ^{2} + x + 1}

f (x) = 7 + 10 x

y = x + 1, y = \frac{x ^{2}}{4}, x = 0, x = 1

y^{^{'}} = \frac{x ^{2} - 3 y ^{2}}{2 x y}

x \to 11 lim (3)

What is the integral of (81)/(x^3-9x^2) ?
The integral of (81)/(x^3-9x^2) is -ln|x|+9/x+ln|x-9|+C