What is the integral of (x^4-5x^3+5x^2+8)/(x^3+3x^2) ?

The integral of (x^4-5x^3+5x^2+8)/(x^3+3x^2) is -11x+269/9 ln|x+3|+1/2 (x+3)^2-8/(3x)-8/9 ln|x|-33+C

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integral of (x^4-5x^3+5x^2+8)/(x^3+3x^2)

\int \frac{x ^{4} - 5 x ^{3} + 5 x ^{2} + 8}{x ^{3} + 3 x ^{2}} d x

- 11 x + \frac{269}{9} ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - \frac{8}{3 x} - \frac{8}{9} ln ∣ x ∣ - 33 + C

Solution steps

\int \frac{x ^{4} - 5 x ^{3} + 5 x ^{2} + 8}{x ^{3} + 3 x ^{2}} d x

Expand

\frac{x ^{4} - 5 x ^{3} + 5 x ^{2} + 8}{x ^{3} + 3 x ^{2}} : \frac{x ^{4}}{x ^{3} + 3 x ^{2}} - \frac{5 x ^{3}}{x ^{3} + 3 x ^{2}} + \frac{5 x ^{2}}{x ^{3} + 3 x ^{2}} + \frac{8}{x ^{3} + 3 x ^{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 ^{4}}{x ^{3} + 3 x ^{2}} d x - \int \frac{5 x ^{3}}{x ^{3} + 3 x ^{2}} d x + \int \frac{5 x ^{2}}{x ^{3} + 3 x ^{2}} d x + \int \frac{8}{x ^{3} + 3 x ^{2}} d x

\int \frac{x ^{4}}{x ^{3} + 3 x ^{2}} d x = \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣

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

\int \frac{5 x ^{2}}{x ^{3} + 3 x ^{2}} d x = 5 ln ∣ x + 3 ∣

\int \frac{8}{x ^{3} + 3 x ^{2}} d x = 8 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

= \frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 5 (x + 3 - 3 ln ∣ x + 3 ∣) + 5 ln ∣ x + 3 ∣ + 8 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

Simplify

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ - 5 (x + 3 - 3 ln ∣ x + 3 ∣) + 5 ln ∣ x + 3 ∣ + 8 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣) : - 11 x + \frac{269}{9} ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - \frac{8}{3 x} - \frac{8}{9} ln ∣ x ∣ - 33

= - 11 x + \frac{269}{9} ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - \frac{8}{3 x} - \frac{8}{9} ln ∣ x ∣ - 33

Add a constant to the solution

= - 11 x + \frac{269}{9} ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - \frac{8}{3 x} - \frac{8}{9} ln ∣ x ∣ - 33 + C

f (x) = 5 x^{2} (x^{3} + 2)

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

\frac{d}{d x} (5 - 2 sin (x) - e^{x})

\int \frac{t}{( 3 t ^{2} + 1 ) ^{4}} d t

y^{^{'}} + y = cos (t), y (0) = 1

What is the integral of (x^4-5x^3+5x^2+8)/(x^3+3x^2) ?
The integral of (x^4-5x^3+5x^2+8)/(x^3+3x^2) is -11x+269/9 ln|x+3|+1/2 (x+3)^2-8/(3x)-8/9 ln|x|-33+C