- #1
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:\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=\)\(\displaystyle \to\)\(\displaystyle \frac{(1-y)^3}{y}+3(1-y)^2\log y-6\left(-\frac{1}{y}+\frac{y^2}{2}\log y-\frac{y^2}{4}\right)=\)\(\displaystyle \frac{(1-y)^3}{y}+3(1-y)^2\log y+\frac{6}{y}-3y^2\log y+\frac{3y^2}{2}\)Also, y is a dummy variable here, much like your x, so the answer above is identical toDreamWeaver said:\(\displaystyle \frac{d}{dy}\frac{1}{(1-y)}=-\frac{1}{(1-y)^2}\)
\(\displaystyle \Rightarrow\)
\(\displaystyle \frac{y^3}{(1-y)}-3\int\frac{y^2}{(1-y)}
\,dy=\)
\(\displaystyle \frac{y^3}{(1-y)}-3\left[-y^2\log(1-y)+2\int y\log(1-y)\,dy\right]=\)
\(\displaystyle \frac{y^3}{(1-y)}+3y^2\log(1-y)-6\int y\log(1-y)\,dy\)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.DreamWeaver said:\(\displaystyle \frac{(1-x)^3}{x}+3(1-x)^2\log x+\frac{6}{x}-3x^2\log x+\frac{3x^2}{2}\)
Does that match your book?
This one is correct.bergausstein said:...$\displaystyle\int xdx-3\int dx+3\int\frac{1}{x}dx-\int\frac{dx}{x^2} = \frac{x^2}{2}-3x+3ln(x)+\frac{1}{x}+C$
The concept of "Integrate Without Expanding Numerator" refers to a technique in calculus where the numerator of a fraction is not expanded into individual terms before integrating. This is commonly used when the numerator contains a product of two or more functions, making it difficult to integrate using traditional methods.
This technique is used to simplify the integration process and make it easier to solve complex integrals. By not expanding the numerator, the integral can often be solved using simpler techniques such as substitution or integration by parts. It can also help avoid mistakes and save time when integrating.
The main benefit of using this technique is that it can simplify the integration process and make it easier to solve complex integrals. It can also help avoid mistakes and save time when integrating. Additionally, it can be used to solve integrals that are not solvable using traditional methods.
While "Integrate Without Expanding Numerator" can be a useful technique, it is not applicable in all cases. It is most effective when the numerator contains a product of two or more functions, but may not work for other types of integrals. Additionally, it may not always result in the most simplified solution.
Sure, for example, when integrating the function ∫(x+1)e^x dx, we can use the technique of "Integrate Without Expanding Numerator". Instead of expanding the numerator (x+1), we can leave it as is and integrate using substitution. The resulting integral is ∫(x+1)e^x dx = ∫ue^u du, which can be easily solved using integration by parts.