What is the integral of (2x)/((x-1)^5) ?

The integral of (2x)/((x-1)^5) is 2(-x/(4(x-1)^4)-1/(12(x-1)^3))+C

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integral of (2x)/((x-1)^5)

\int \frac{2 x}{( x - 1 ) ^{5}} d x

2 (- \frac{x}{4 ( x - 1 ) ^{4}} - \frac{1}{12 ( x - 1 ) ^{3}}) + C

Solution steps

\int \frac{2 x}{( x - 1 ) ^{5}} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{( x - 1 ) ^{5}} d x

Apply Integration By Parts

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

= 2 (- \frac{x}{4 ( x - 1 ) ^{4}} - \int - \frac{1}{4 ( x - 1 ) ^{4}} d x)

\int - \frac{1}{4 ( x - 1 ) ^{4}} d x = \frac{1}{12 ( x - 1 ) ^{3}}

= 2 (- \frac{x}{4 ( x - 1 ) ^{4}} - \frac{1}{12 ( x - 1 ) ^{3}})

Add a constant to the solution

= 2 (- \frac{x}{4 ( x - 1 ) ^{4}} - \frac{1}{12 ( x - 1 ) ^{3}}) + C

\frac{d}{d x} (\frac{3}{1 - x})

f (t) = ln (t)

\int 2^{s i n (x)} cos (x) d x

0.145 \cdot \frac{d v}{d t} = (0.145 \cdot 9.81) - (0.043 \cdot v)

(x, y) \to (0, 0) lim (∣ x + y ∣)

What is the integral of (2x)/((x-1)^5) ?
The integral of (2x)/((x-1)^5) is 2(-x/(4(x-1)^4)-1/(12(x-1)^3))+C