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paulmdrdo
Active member
- May 13, 2013
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how to solve this without expanding the numerator.
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$
how to solve this without expanding the numerator.
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$
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 $y=1-x$.no, your answer doesn't match the answer in my book.
no, your answer doesn't match the answer in my book.
\(\displaystyle \frac{d}{dy}\frac{1}{(1-y)}=-\frac{1}{(1-y)^2}\)
\(\displaystyle \Rightarrow\)
\(\displaystyle \frac{y^3}{(1-y)}-3\int\frac{y^2}{(1-y)}
\,dy=\)
\(\displaystyle \frac{y^3}{(1-y)}-3\left[-y^2\log(1-y)+2\int y\log(1-y)\,dy\right]=\)
\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\int y\log(1-y)\,dy\)
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
\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=\)
\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)+\frac{6}{y}-3y^2\log y+\frac{3y^2}{2}\)
I don't know what the book says, but yours is not correct.\(\displaystyle \frac{(1-x)^3}{x}+3(1-x)^2\log x+\frac{6}{x}-3x^2\log x+\frac{3x^2}{2}\)
Does that match your book???
This one is correct....$\displaystyle\int xdx-3\int dx+3\int\frac{1}{x}dx-\int\frac{dx}{x^2} = \frac{x^2}{2}-3x+3ln(x)+\frac{1}{x}+C$