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RolyPoly
Group Title
This is the first dish
\[\int \frac{x^2}{1x^2}dx\]
Apart from resolving that into partial fractions, what is a better way to do it?
 2 years ago
 2 years ago
RolyPoly Group Title
This is the first dish \[\int \frac{x^2}{1x^2}dx\] Apart from resolving that into partial fractions, what is a better way to do it?
 2 years ago
 2 years ago

This Question is Closed

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
let x=sinu \[\int \frac{x^2}{1x^2}dx = \int\frac{sin^2ucosu}{cos^2u}du=\int(sec^2ucosu)du=tanusinu+C\]?
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Last line there: tanu sinu +C \[=\frac{x}{x^2+1}x+C\] What am I doing here :S
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{sin^2ucosu}{cos^2u}=\frac{(1cos^2u)cosu}{cos^2u} = sec^2ucosu\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
tan u is x/ sqrt(1x^2)
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Oh.. My mistake! \[=\frac{x}{1x^2} x+C\]
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Alright, let's go to the main dish :S
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
and its not sec^2 , check again
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
secu! I'd better do it all over again!!! \[\int \frac{x^2}{1x^2}dx \]\[= \int\frac{sin^2ucosu}{cos^2u}du\]\[=\int(secucosu)du\]\[=\lnsecu+tanusinu+C\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
not a better way......
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I know, I just try whatever I can.. Is there a better way then?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
i don't think,partial is best....
 2 years ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
\[\int \frac{x^2}{1x^2}dx\]\[ = \int (\frac{1}{2(x+1)}\frac{1}{2(x1)}1)dx\]\[=\frac{1}{2} lnx+1 \frac{1}{2}lnx1x+C\]\[=\frac{1}{2}ln\frac{x+1}{x1}x+C\]
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
yup, thats correct.....
 2 years ago
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