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

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

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

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

2 x + \frac{1}{2} (x - 1)^{2} + 3 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ + \frac{2}{x} - 2 + C

Solution steps

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

Expand

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

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

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

= \frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣ + 2 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

Simplify

\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣ + 2 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣) : 2 x + \frac{1}{2} (x - 1)^{2} + 3 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ + \frac{2}{x} - 2

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

Add a constant to the solution

= 2 x + \frac{1}{2} (x - 1)^{2} + 3 ln ∣ x - 1 ∣ - 2 ln ∣ x ∣ + \frac{2}{x} - 2 + C

\frac{d}{d x} (4 π^{2})

- 4 sin (t)

n = 1 \sum \infty (\frac{1}{2})^{2 n}

x^{2} y^{^{'}} - 3 x y = sin (2 x) x^{5}

x \to 2 + lim (\frac{x ^{2}}{4 - x ^{2}})

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