Question

How to solve this integral:

Answer 1

\[\int\limits_{}^{}1/(1-x ^{2})\]

Answer 2

what does that long thing before the first one mean?

Answer 3

hint:x=sinu....dx=cosu.du

Answer 4

ln(x+1/1-x^2)

Answer 5

i still don't get it

Answer 6

SKforlife: It's the symbol for integral equations :D

Answer 7

integ_(cosu.du/1-sin^2(u))=integ_cosudu/cos^2(u)=integ_du/cosu=integ_secu.du
integ_secu.du=ln(secu+tanu) draw a triangle..your angle is u..sinu=x then write each other.

Answer 8

i\[=\int\limits{\frac{1}{(1-x)(1+x)}dx}\]break intp partial fractions\[\frac{1}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x}\]

Answer 9

\[\frac{1}{1-x^2}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\]

Answer 10

yeah what nikvist did xD

Answer 11

Did u understnd?

Answer 12

yes, thanks

Answer 13

:)