What is the integral of (x^2-64)/(x(x-1)^2) ?

The integral of (x^2-64)/(x(x-1)^2) is 65ln|x-1|+(63)/(x-1)-64ln|x|+C

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integral of (x^2-64)/(x(x-1)^2)

\int \frac{x ^{2} - 64}{x ( x - 1 ) ^{2}} d x

65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{2} - 64}{x ( x - 1 ) ^{2}} d x

Expand

\frac{x ^{2} - 64}{x ( x - 1 ) ^{2}} : \frac{x ^{2}}{x ( x - 1 ) ^{2}} - \frac{64}{x ( x - 1 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{2}}{x ( x - 1 ) ^{2}} d x - \int \frac{64}{x ( x - 1 ) ^{2}} d x

\int \frac{x ^{2}}{x ( x - 1 ) ^{2}} d x = ln ∣ x - 1 ∣ - \frac{1}{x - 1}

\int \frac{64}{x ( x - 1 ) ^{2}} d x = 64 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1})

= ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 64 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1})

Simplify

ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 64 (ln ∣ x ∣ - ln ∣ x - 1 ∣ - \frac{1}{x - 1}) : 65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣

= 65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣

Add a constant to the solution

= 65 ln ∣ x - 1 ∣ + \frac{63}{x - 1} - 64 ln ∣ x ∣ + C

\int \frac{1}{y 1 + ( ln ( y ) ) ^{2}} d y

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

\int sec (t) (8 sec (t) + 5 tan (t)) d t

\frac{d}{d x} ((2 x)^{3})

x \to 0 lim ((x)^{3})

What is the integral of (x^2-64)/(x(x-1)^2) ?
The integral of (x^2-64)/(x(x-1)^2) is 65ln|x-1|+(63)/(x-1)-64ln|x|+C