What is the integral of (3x^2-21x+32)/(x^3-8x^2+16x) ?

The integral of (3x^2-21x+32)/(x^3-8x^2+16x) is ln|x-4|+1/(x-4)+2ln|x|+C

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integral of (3x^2-21x+32)/(x^3-8x^2+16x)

\int \frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} d x

ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} d x

Expand

\frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} : \frac{3 x ^{2}}{x ^{3} - 8 x ^{2} + 16 x} - \frac{21 x}{x ^{3} - 8 x ^{2} + 16 x} + \frac{32}{x ^{3} - 8 x ^{2} + 16 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{3 x ^{2}}{x ^{3} - 8 x ^{2} + 16 x} d x - \int \frac{21 x}{x ^{3} - 8 x ^{2} + 16 x} d x + \int \frac{32}{x ^{3} - 8 x ^{2} + 16 x} d x

\int \frac{3 x ^{2}}{x ^{3} - 8 x ^{2} + 16 x} d x = 3 (ln ∣ x - 4 ∣ - \frac{4}{x - 4})

\int \frac{21 x}{x ^{3} - 8 x ^{2} + 16 x} d x = - \frac{21}{x - 4}

\int \frac{32}{x ^{3} - 8 x ^{2} + 16 x} d x = 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

= 3 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - (- \frac{21}{x - 4}) + 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

Simplify

3 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - (- \frac{21}{x - 4}) + 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )}) : ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣

= ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣

Add a constant to the solution

= ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣ + C

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

\frac{\partial}{\partial w} ((z + w) x^{\frac{1}{2}})

f (x) = sin (\frac{1}{x})

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 6

\frac{\partial}{\partial x} (4 x cos (x) cos (y))

What is the integral of (3x^2-21x+32)/(x^3-8x^2+16x) ?
The integral of (3x^2-21x+32)/(x^3-8x^2+16x) is ln|x-4|+1/(x-4)+2ln|x|+C