Question

Integral of 1/(1-y2) Integral of 1/(1-y2) @Mathematics

Answer 1

\[\int\limits y/(1+y)^2\]
This is what I'm really looking to solve

Answer 2

\[\int\limits y/\sqrt{(1+y^2)}\]

Answer 3

\[\frac{ 1 }{ 1-y^2 }=\frac{ 1 }{ (1+y)(1-y) }=\frac{ 1 }{ 1+y }+\frac{ 1 }{ 1-y }\]

\[\int\limits(\frac{ 1 }{ 1+y }+\frac{ 1 }{ 1-y })dy=\ln|1+y|-\ln|1-y|=\ln|\frac{ 1+y }{ 1-y }|\]

Answer 4

Ah yes, that makes sense

Answer 5

But I mistyped my equation, unfortunately.

Answer 6

let
\[u=1+y^2\]\[du=2y dy\]\[ydy=\frac{ 1 }{ 2 }du\]
\[\int\limits \frac{ ydy }{ \sqrt{1+y^2} }=\int\limits \frac{ du }{ u^\frac{ 1 }{ 2 } }\]\[=-\frac{ 1 }{ 2 }u^\frac{ -3 }{ 2 }\]\[=-\frac{ 1 }{ 2 }(1+y^2)^\frac{ -3 }{ 2 }\]

Answer 7

LOL the last 2 lines are wrong ahah

Answer 8

should be:\[=2u^ \frac{ 1 }{ 2 }\]\[=2(1+y^2)^\frac{ 1 }{ 2 }\]

Answer 9

Thank you!

Answer 10

sin(x+y)dy=dx