What is the integral of 7x^3(x-1)^{-4} ?

The integral of 7x^3(x-1)^{-4} is 7(ln|x-1|-3/(x-1)-3/(2(x-1)^2)-1/(3(x-1)^3))+C

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integral of 7x^3(x-1)^{-4}

\int 7 x^{3} (x - 1)^{- 4} d x

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

Solution steps

\int 7 x^{3} (x - 1)^{- 4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int x^{3} (x - 1)^{- 4} d x

Apply u-substitution

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

= 7 \cdot \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}}

= 7 \cdot \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

= 7 (\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}}

= 7 (ln ∣ u ∣ - \frac{3}{u} - \frac{3}{2 u ^{2}} - \frac{1}{3 u ^{3}})

Substitute back

u = x - 1

= 7 (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

= 7 (ln ∣ x - 1 ∣ - \frac{3}{x - 1} - \frac{3}{2 ( x - 1 ) ^{2}} - \frac{1}{3 ( x - 1 ) ^{3}}) + C

y = \frac{a - x}{a + x}

\frac{d y}{d x} - \frac{y}{x} = \frac{x}{y}

x, \frac{1}{2} x, x = 9

\frac{\partial}{\partial x} (- x e^{3 x y})

\int_{- 1}^{5} (3 e^{- x})^{2} d x

What is the integral of 7x^3(x-1)^{-4} ?
The integral of 7x^3(x-1)^{-4} is 7(ln|x-1|-3/(x-1)-3/(2(x-1)^2)-1/(3(x-1)^3))+C