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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let x=sinu \[\int \frac{x^2}{1x^2}dx = \int\frac{sin^2ucosu}{cos^2u}du=\int(sec^2ucosu)du=tanusinu+C\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Last line there: tanu sinu +C \[=\frac{x}{x^2+1}x+C\] What am I doing here :S

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{sin^2ucosu}{cos^2u}=\frac{(1cos^2u)cosu}{cos^2u} = sec^2ucosu\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh.. My mistake! \[=\frac{x}{1x^2} x+C\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright, let's go to the main dish :S

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0secu! I'd better do it all over again!!! \[\int \frac{x^2}{1x^2}dx \]\[= \int\frac{sin^2ucosu}{cos^2u}du\]\[=\int(secucosu)du\]\[=\lnsecu+tanusinu+C\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know, I just try whatever I can.. Is there a better way then?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{x^2}{1x^2}dx\]\[ = \int (\frac{1}{2(x+1)}\frac{1}{2(x1)}1)dx\]\[=\frac{1}{2} lnx+1 \frac{1}{2}lnx1x+C\]\[=\frac{1}{2}ln\frac{x+1}{x1}x+C\]
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