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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

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

Simplify

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

Add a constant to the solution

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

cos (3 x)

\frac{d}{d x} (\frac{10}{ln ( x )})

t^{2} \frac{d y}{d t} + y = 0, y (7) = - 4

\int sin (2 t) (cot (2 t) + 1) d t

\frac{d}{d x} (∣ x - 1 ∣^{3})

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