What is the integral from 1 to 2 of x/((1-x)^4) ?

The integral from 1 to 2 of x/((1-x)^4) is diverges

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integral from 1 to 2 of x/((1-x)^4)

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

diverges

Solution steps

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

Take the partial fraction of

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

= \int_{1}^{2} \frac{1}{( x - 1 ) ^{3}} + \frac{1}{( x - 1 ) ^{4}} 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_{1}^{2} \frac{1}{( x - 1 ) ^{3}} d x + \int_{1}^{2} \frac{1}{( x - 1 ) ^{4}} d x

\int_{1}^{2} \frac{1}{( x - 1 ) ^{3}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 1 to 2 of x/((1-x)^4) ?
The integral from 1 to 2 of x/((1-x)^4) is diverges