Question

\[evaluated the interval \int\limits_{}^{} x/(x-1)^{3}dx\]

Answer 1

\[\frac{x}{(x-1)^3}=\frac{1}{(x-
1)^2}+\frac{1}{(x-1)^3}

\]

By Partial fraction, then it is easy.

Answer 2

\[
\int \frac{1}{(x-1)^3} \, dx = -\frac{1}{2 (x-1)^2} +C
\]

Answer 3

\[
\int \frac{1}{(x-1)^2} \, dx =-\frac{1}{x-1} + C
\]

Answer 4

Combining the tw and simplifying, you get
the answer as
\[ \frac{1-2 x}{2 (x-1)^2} + C

\]

Answer 5

That is what we combined to get the final answer
\[-\frac{1}{x-1}-\frac{1}{2
(x-1)^2}

\]

Answer 6

Use u-substitution

take \[u=x-1\]
then \[du=dx\]

This simplifies the the denominator.
Now substitute u in....u=x-1 and x=u+1

Now you have: \[\int\limits_{}^{} ((u+1)/u^{3})du\]
(I can't figure out how to show my division as a fraction on this site yet, sorry)

Now break up your fraction into two different ones

\[\int\limits_{}^{} ((u/u^{3})+(1/u ^{3}))du\]

Simplify: \[\int\limits_{}^{} ((1/u^{2})+(1/u ^{3}))du\]

Now you can take the integral of the two fractions individually.

\[\int\limits_{}^{} (1/u^2)du=-1/u+c_{1}\]

\[\int\limits_{}^{} (1/u^3)du=-1/(2u^2)+c_{2}\]

Don't forget to add the two fractions:
\[-1/u+-1/(2u^2)+c\]

Simplify (common denominators): \[(-2u-1)/(2u^2)+c\]
Substitute x-1 back in for u
\[(-2(x-1)-1)/(2(x-1)^2)\]
\[(-2x+2-1)/(2(x-1)^2)\]
\[(-2x+1)/(2(x-1)^2)\]

(Again, sorry about the notation. How do you represent division as a fraction using this site?)

Answer 7

to do fractions use \frac{a}{b}

\[\frac{a}{b}\]