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

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

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

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

- 4 ln ∣ x ∣ + 5 ln ∣ x - 1 ∣ + \frac{3}{x - 1} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} - 4}{x ^{3} - 2 x ^{2} + x} : - \frac{4}{x} + \frac{5}{x - 1} - \frac{3}{( x - 1 ) ^{2}}

= \int - \frac{4}{x} + \frac{5}{x - 1} - \frac{3}{( x - 1 ) ^{2}} 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

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int \frac{5}{x - 1} d x = 5 ln ∣ x - 1 ∣

\int \frac{3}{( x - 1 ) ^{2}} d x = - \frac{3}{x - 1}

= - 4 ln ∣ x ∣ + 5 ln ∣ x - 1 ∣ - (- \frac{3}{x - 1})

Simplify

= - 4 ln ∣ x ∣ + 5 ln ∣ x - 1 ∣ + \frac{3}{x - 1}

Add a constant to the solution

= - 4 ln ∣ x ∣ + 5 ln ∣ x - 1 ∣ + \frac{3}{x - 1} + C

\frac{d}{d x} (2^{x s i n (x)})

f (x) = 5 tan (x)

\frac{d}{d x} (e cos (x))

\int e^{t} cos (e^{t}) d t

f (x) = - 5

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