What is the integral of (2x^3+11x+8)/(x^3+4x^2+4x) ?

The integral of (2x^3+11x+8)/(x^3+4x^2+4x) is 2x-10ln|x+2|-(15)/(x+2)+2ln|x|+4+C

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integral of (2x^3+11x+8)/(x^3+4x^2+4x)

\int \frac{2 x ^{3} + 11 x + 8}{x ^{3} + 4 x ^{2} + 4 x} d x

2 x - 10 ln ∣ x + 2 ∣ - \frac{15}{x + 2} + 2 ln ∣ x ∣ + 4 + C

Solution steps

\int \frac{2 x ^{3} + 11 x + 8}{x ^{3} + 4 x ^{2} + 4 x} d x

Factor

\frac{2 x ^{3} + 11 x + 8}{x ^{3} + 4 x ^{2} + 4 x}

= \int \frac{2 x ^{3} + 11 x + 8}{x ( x + 2 ) ^{2}} d x

Expand

\frac{2 x ^{3} + 11 x + 8}{x ( x + 2 ) ^{2}} : \frac{2 x ^{3}}{x ( x + 2 ) ^{2}} + \frac{11 x}{x ( x + 2 ) ^{2}} + \frac{8}{x ( x + 2 ) ^{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{2 x ^{3}}{x ( x + 2 ) ^{2}} d x + \int \frac{11 x}{x ( x + 2 ) ^{2}} d x + \int \frac{8}{x ( x + 2 ) ^{2}} d x

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

\int \frac{11 x}{x ( x + 2 ) ^{2}} d x = - \frac{11}{x + 2}

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

= 2 (x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2}) - \frac{11}{x + 2} + 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )})

Simplify

2 (x + 2 - 4 ln ∣ x + 2 ∣ - \frac{4}{x + 2}) - \frac{11}{x + 2} + 8 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 2 ∣ + \frac{1}{2 ( x + 2 )}) : 2 x - 10 ln ∣ x + 2 ∣ - \frac{15}{x + 2} + 2 ln ∣ x ∣ + 4

= 2 x - 10 ln ∣ x + 2 ∣ - \frac{15}{x + 2} + 2 ln ∣ x ∣ + 4

Add a constant to the solution

= 2 x - 10 ln ∣ x + 2 ∣ - \frac{15}{x + 2} + 2 ln ∣ x ∣ + 4 + C

w \to 0 lim (\frac{a - w ^{2} + a ^{2}}{b - w ^{2} + b ^{2}})

t an g e n t 2 x^{2} + x - 3, at x = - 1

\int \frac{z}{1 - y ^{2} z ^{2}} d y

d er i v a t i v e y = \frac{x}{7} - \frac{7}{x}

x \to 0 lim (x ln (x) - x)

What is the integral of (2x^3+11x+8)/(x^3+4x^2+4x) ?
The integral of (2x^3+11x+8)/(x^3+4x^2+4x) is 2x-10ln|x+2|-(15)/(x+2)+2ln|x|+4+C