paulmdrdo1
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how to solve this without expanding the numerator.
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$
Not sure if this counts as cheating or not, but I'd say contract the numerator instead! (Bandit)paulmdrdo said:how to solve this without expanding the numerator.
$\displaystyle \int\frac{(x-1)^3}{x^2}dx$
paulmdrdo said: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 said:no, your answer doesn't match the answer in my book.
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:$$\frac{y^3}{(1-y)}+3y^2\log(1-y)-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=$$$$\to$$$$\frac{(1-y)^3}{y}+3(1-y)^2\log y-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=$$$$\frac{(1-y)^3}{y}+3(1-y)^2\log y+\frac{6}{y}-3y^2\log y+\frac{3y^2}{2}$$Also, y is a dummy variable here, much like your x, so the answer above is identical toDreamWeaver said:$$\frac{d}{dy}\frac{1}{(1-y)}=-\frac{1}{(1-y)^2}$$
$$\Rightarrow$$
$$\frac{y^3}{(1-y)}-3\int\frac{y^2}{(1-y)}
\,dy=$$
$$\frac{y^3}{(1-y)}-3\left[-y^2\log(1-y)+2\int y\log(1-y)\,dy\right]=$$
$$\frac{y^3}{(1-y)}+3y^2\log(1-y)-6\int y\log(1-y)\,dy$$For that last integral, use the same substitution as before to obtain
$$\int y\log(1-y)\,dy=-\int (1-y)\log y\,dy=-\frac{1}{y}+\int y\log y\,dy= $$
$$-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{1}{2}\int y\,dy=$$
$$-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}$$
The final answer is then
$$\frac{y^3}{(1-y)}+3y^2\log(1-y)-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=$$
$$\frac{y^3}{(1-y)}+3y^2\log(1-y)+\frac{6}{y}-3y^2\log y+\frac{3y^2}{2}$$
I don't know what the book says, but yours is not correct.DreamWeaver said:$$\frac{(1-x)^3}{x}+3(1-x)^2\log x+\frac{6}{x}-3x^2\log x+\frac{3x^2}{2}$$
Does that match your book?
This one is correct.bergausstein said:...$\displaystyle\int xdx-3\int dx+3\int\frac{1}{x}dx-\int\frac{dx}{x^2} = \frac{x^2}{2}-3x+3ln(x)+\frac{1}{x}+C$