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

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

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integral of (x^4+1)/(x^3-x^2)

\int \frac{x ^{4} + 1}{x ^{3} - x ^{2}} d x

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

Solution steps

\int \frac{x ^{4} + 1}{x ^{3} - x ^{2}} d x

Long division

\frac{x ^{4} + 1}{x ^{3} - x ^{2}} : x + 1 + \frac{x ^{2} + 1}{x ^{3} - x ^{2}}

= \int x + 1 + \frac{x ^{2} + 1}{x ^{3} - x ^{2}} d 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 x d x + \int 1 d x + \int \frac{x ^{2} + 1}{x ^{3} - x ^{2}} d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

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

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

Add a constant to the solution

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

\int_{7}^{9} \frac{y}{y ^{2} - 3 y - 4} d y

\frac{d}{d t} (e^{4 t} sin (t))

f (x) = 3 x^{2} - x^{3}

\int (9 + u + u^{2}) d u

\int \frac{4}{x ^{2} ( x - 2 )} d x

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