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Integrate Without Expanding Numerator

paulmdrdo

Active member
May 13, 2013
386
how to solve this without expanding the numerator.
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$
 

DreamWeaver

Well-known member
Sep 16, 2013
337
how to solve this without expanding the numerator.
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$

Not sure if this counts as cheating or not, but I'd say contract the numerator instead!! (Bandit)

Substitute \(\displaystyle y=1-x\,\) \(\displaystyle \Rightarrow\)


\(\displaystyle \int\frac{(x-1)^3}{x^2}\,dx=-\int\frac{y^3}{(1-y)^2}\,dy\)


Full solution...


\(\displaystyle \frac{d}{dy}\frac{1}{(1-y)}=-\frac{1}{(1-y)^2}\)

\(\displaystyle \Rightarrow\)

\(\displaystyle \frac{y^3}{(1-y)}-3\int\frac{y^2}{(1-y)}
\,dy=\)

\(\displaystyle \frac{y^3}{(1-y)}-3\left[-y^2\log(1-y)+2\int y\log(1-y)\,dy\right]=\)

\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\int y\log(1-y)\,dy\)


For that last integral, use the same substitution as before to obtain

\(\displaystyle \int y\log(1-y)\,dy=-\int (1-y)\log y\,dy=-\frac{1}{y}+\int y\log y\,dy= \)

\(\displaystyle -\frac{1}{y}+\frac{y^2}{2}\log y-\frac{1}{2}\int y\,dy=\)

\(\displaystyle -\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\)

The final answer is then

\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=\)

\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)+\frac{6}{y}-3y^2\log y+\frac{3y^2}{2}\)
 

paulmdrdo

Active member
May 13, 2013
386
no, your answer doesn't match the answer in my book.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
no, your answer doesn't match the answer in my book.
I haven't checked this myself but did you remember to back-substitute $x$ in for $y$? His final answer is in terms of $y$ but he made the initial substitution $y=1-x$.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
no, your answer doesn't match the answer in my book.

However, that doesn't make the one above incorrect. What form is given in your book? ;)
 

paulmdrdo

Active member
May 13, 2013
386
the answer is in terms of x when expanding the numerator.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
OK, so here was my solution...


\(\displaystyle \frac{d}{dy}\frac{1}{(1-y)}=-\frac{1}{(1-y)^2}\)

\(\displaystyle \Rightarrow\)

\(\displaystyle \frac{y^3}{(1-y)}-3\int\frac{y^2}{(1-y)}
\,dy=\)

\(\displaystyle \frac{y^3}{(1-y)}-3\left[-y^2\log(1-y)+2\int y\log(1-y)\,dy\right]=\)

\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\int y\log(1-y)\,dy\)


For that last integral, use the same substitution as before to obtain

\(\displaystyle \int y\log(1-y)\,dy=-\int (1-y)\log y\,dy=-\frac{1}{y}+\int y\log y\,dy= \)

\(\displaystyle -\frac{1}{y}+\frac{y^2}{2}\log y-\frac{1}{2}\int y\,dy=\)

\(\displaystyle -\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\)

The final answer is then

\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=\)

\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)+\frac{6}{y}-3y^2\log y+\frac{3y^2}{2}\)

In the line just above that last one, the terms on the RHS encased in large brackets have already been back-substituted, so now I'll do the other bit:


\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=\)


\(\displaystyle \to\)


\(\displaystyle \frac{(1-y)^3}{y}+3(1-y)^2\log y-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=\)


\(\displaystyle \frac{(1-y)^3}{y}+3(1-y)^2\log y+\frac{6}{y}-3y^2\log y+\frac{3y^2}{2}\)


Also, y is a dummy variable here, much like your x, so the answer above is identical to

\(\displaystyle \frac{(1-x)^3}{x}+3(1-x)^2\log x+\frac{6}{x}-3x^2\log x+\frac{3x^2}{2}\)

Does that match your book???
 

bergausstein

Active member
Jul 30, 2013
191
i tried to solve this by expanding the numerator of the integrand,

$\displaystyle\int\frac{(x-1)^3}{x^2}dx = \int \frac{(x^3-3x^2+3x-1)}{x^2}dx = \int\frac{x^3}{x^2}dx-3\int\frac{x^2}{x^2}dx+3\int\frac{x}{x^2}dx-\int\frac{dx}{x^2}$...

...$\displaystyle\int xdx-3\int dx+3\int\frac{1}{x}dx-\int\frac{dx}{x^2} = \frac{x^2}{2}-3x+3ln(x)+\frac{1}{x}+C$
 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
\(\displaystyle \frac{(1-x)^3}{x}+3(1-x)^2\log x+\frac{6}{x}-3x^2\log x+\frac{3x^2}{2}\)

Does that match your book???
I don't know what the book says, but yours is not correct.

...$\displaystyle\int xdx-3\int dx+3\int\frac{1}{x}dx-\int\frac{dx}{x^2} = \frac{x^2}{2}-3x+3ln(x)+\frac{1}{x}+C$
This one is correct.

-Dan
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
I decided to just keep applying integration by parts until it wasn't needed anymore (i.e. three times):

First time: $u=(x-1)^3$, $dv=\dfrac{1}{x^2}\,dx \implies \,du=3(x-1)^2\,dx$ and $v=-\dfrac{1}{x}$.

Thus,

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

Second time: $u=(x-1)^2$, $dv = \dfrac{1}{x}\,dx\implies \,du=2(x-1)\,dx = (2x-2) \,dx$ and $v=\ln x$.

Thus,

\[\color{red}{\int\frac{(x-1)^2}{x}\,dx} = (x-1)^2\ln x - \int (2x-2)\ln x\,dx\]

and hence

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

Third time: $u=\ln x$, $dv = (2x-2)\,dx \implies \,du=\dfrac{1}{x}\,dx$ and $v=x^2-2x$.

Thus,

\[\begin{aligned}\color{blue}{\int(2x-2)\ln x \,dx} &= (x^2-2x)\ln x - \int \frac{x^2-2x}{x}\,dx\\ &= (x^2-2x)\ln x - \int x-2 \,dx \\ &= (x^2-2x)\ln x - \frac{x^2}{2} + 2x + C\end{aligned}\]

Therefore, by not expanding the numerator, we get

\[\int\frac{(x-1)^3}{x^2}\,dx = -\frac{(x-1)^3}{x} + 3(x-1)^2\ln x - 3(x^2-2x)\ln x + \frac{3}{2}x^2 -6x+C\]