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

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

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integral of x^3(x-1)^{-4}

\int x^{3} (x - 1)^{- 4} d x

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

Solution steps

\int x^{3} (x - 1)^{- 4} d x

Apply u-substitution

: \int \frac{( u + 1 ) ^{3}}{u ^{4}} d u

= \int \frac{( u + 1 ) ^{3}}{u ^{4}} d u

Expand

\frac{( u + 1 ) ^{3}}{u ^{4}} : \frac{1}{u} + \frac{3}{u ^{2}} + \frac{3}{u ^{3}} + \frac{1}{u ^{4}}

= \int \frac{1}{u} + \frac{3}{u ^{2}} + \frac{3}{u ^{3}} + \frac{1}{u ^{4}} d u

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{3}{u ^{2}} d u = - \frac{3}{u}

\int \frac{3}{u ^{3}} d u = - \frac{3}{2 u ^{2}}

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

= ln ∣ u ∣ - \frac{3}{u} - \frac{3}{2 u ^{2}} - \frac{1}{3 u ^{3}}

Substitute back

u = x - 1

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

Add a constant to the solution

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

\frac{s ^{4}}{s}

f (x) = 3 x^{2} + 3 x + 4

- x^{2} + 6 x

\frac{d}{d x} (\frac{8}{1 + x})

f (x) = ln (tan (x))

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