Integral - what is wrong here?

[Shinta]

Hi the problem is to integrate x^3/(x-1) dx. I know the solution is x^3/3+x^2/2+x+ln(x-1)+constant; I figured out through long division.
I am trying to see why this u-substitution method is not working and provides a different answer. Thanks!

x^3*(x-1)^-1 dx
u=x-1 ; dx=du ; x=u+1
x^3*u^-1 du
substitute u+1 in for x
(u+1)^3*u^-1 du
foil out and divide out u
u^2 + 3u + 3 + 1/u du
integrate
u^3/3 + 3u^2/2 + 3u + ln(u) + C
plug in x-1 for u
(x-1)^3/3 + 3(x-1)^2/2 + 3(x-1) + ln(x-1) +C

 

[Deleted member 4993]

Hi the problem is to integrate x^3/(x-1) dx. I know the solution is x^3/3+x^2/2+x+ln(x-1)+constant; I figured out through long division.
I am trying to see why this u-substitution method is not working and provides a different answer. Thanks!

x^3*(x-1)^-1 dx
u=x-1 ; dx=du ; x=u+1
x^3*u^-1 du
substitute u+1 in for x
(u+1)^3*u^-1 du
foil out and divide out u
u^2 + 3u + 3 + 1/u du
integrate
u^3/3 + 3u^2/2 + 3u + ln(u) + C
plug in x-1 for u
(x-1)^3/3 + 3(x-1)^2/2 + 3(x-1) + ln(x-1) +C

Your solution and the solution given to you are both correct.

If you expand your solution, you'll see that it is same as the other, except for a constant term.

 

[lookagain]

integrate
u^3/3 + 3u^2/2 + 3u + ln(u) + C

plug in x-1 for u

(x-1)^3/3 + 3(x-1)^2/2 + 3(x-1) + ln(x-1) +C

Your instructor, textbook, hand-out, and/or computer programs need to show that
there are absolute value bars when used with the natural log in these particular
antiderivatives:


\(\displaystyle \dfrac{u^3}{3} \ + \ \dfrac{3u^2}{2} \ + \ 3u \ + \ ln|u| \ + C\)


\(\displaystyle \dfrac{(x - 1)^3}{3} \ + \ \dfrac{3(x - 1)^2}{2} \ + \ 3(x - 1) \ + \ ln|x - 1| \ + C\)



- - -- - - - - - - - - - --- - - -

x^3*(x-1)^-1 dx
u=x-1 ; dx=du ; x=u+1\(\displaystyle \ \ \ \ \) <-------

And the logical progression here would be: \(\displaystyle \ \ \)u = x - 1 \(\displaystyle \ \) leads to \(\displaystyle \ \) x = u + 1 \(\displaystyle \ \) leads to \(\displaystyle \ \) dx = du

 

[HallsofIvy]

What I would have done, right from the beginning, is start by doing the indicated division: \(\displaystyle \frac{x}{x- 1}= \frac{x- 1}{x- 1}+ \frac{1}{x- 1}= 1+ \frac{1}{x- 1}\) and integrate that.

 

[lookagain]

What I would have done, right from the beginning, is start by doing the indicated division:

\(\displaystyle \frac{x}{x- 1}= \frac{x- 1}{x- 1}+ \frac{1}{x- 1}= 1+ \frac{1}{x- 1}\) and integrate that.

The original numerator is \(\displaystyle \ x^3, \ \) as opposed to just \(\displaystyle \ x.\)

 

[HallsofIvy]

No wonder I keep failing math courses!

 

[Deleted member 4993]

\(\displaystyle \dfrac{x^3}{x-1} \ = \ \dfrac{x^3 - 1 + 1}{x-1} \ = \ x^2 + x + 1 + \dfrac{1}{x-1} \)