What is the integral of (3x^2+17x+32)/(x^3+8x^2+16x) ?

The integral of (3x^2+17x+32)/(x^3+8x^2+16x) is ln|x+4|+3/(x+4)+2ln|x|+C

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integral of (3x^2+17x+32)/(x^3+8x^2+16x)

\int \frac{3 x ^{2} + 17 x + 32}{x ^{3} + 8 x ^{2} + 16 x} d x

ln ∣ x + 4 ∣ + \frac{3}{x + 4} + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{3 x ^{2} + 17 x + 32}{x ^{3} + 8 x ^{2} + 16 x} d x

Expand

\frac{3 x ^{2} + 17 x + 32}{x ^{3} + 8 x ^{2} + 16 x} : \frac{3 x ^{2}}{x ^{3} + 8 x ^{2} + 16 x} + \frac{17 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{17 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{17 x}{x ^{3} + 8 x ^{2} + 16 x} d x = - \frac{17}{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{17}{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{17}{x + 4} + 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x + 4 ∣ + \frac{1}{4 ( x + 4 )}) : ln ∣ x + 4 ∣ + \frac{3}{x + 4} + 2 ln ∣ x ∣

= ln ∣ x + 4 ∣ + \frac{3}{x + 4} + 2 ln ∣ x ∣

Add a constant to the solution

= ln ∣ x + 4 ∣ + \frac{3}{x + 4} + 2 ln ∣ x ∣ + C

\int (x + 1) 3 x + 2 d x

\frac{d ^{4}}{d x ^{4}} (11 e^{x^{2}})

\frac{d}{d x} (2 x^{3} - 7 x^{2} + 3 x - 2)

\frac{d}{d x} ((7 + 2 x) e^{- 3 x})

t^{2} y^{^{''}} + 2 t y^{^{'}} - 1 = 0

What is the integral of (3x^2+17x+32)/(x^3+8x^2+16x) ?
The integral of (3x^2+17x+32)/(x^3+8x^2+16x) is ln|x+4|+3/(x+4)+2ln|x|+C