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

The integral of y/((y-1)^4) is -1/(2(y-1)^2)-1/(3(y-1)^3)+C

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integral of y/((y-1)^4)

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

- \frac{1}{2 ( y - 1 ) ^{2}} - \frac{1}{3 ( y - 1 ) ^{3}} + C

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u + 1}{u ^{4}} : \frac{1}{u ^{3}} + \frac{1}{u ^{4}}

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

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

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

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

Substitute back

u = y - 1

= - \frac{1}{2 ( y - 1 ) ^{2}} - \frac{1}{3 ( y - 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{2 ( y - 1 ) ^{2}} - \frac{1}{3 ( y - 1 ) ^{3}} + C

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