Question

Help integrate please :
∫ 1/√(1+e^(-2x))

Answer 1

\[I=\int\limits \frac{ dx }{ \sqrt{1-e ^{-2 x}} }\]
\[=\int\limits \frac{ e^x }{\sqrt{e ^{2x}-1} }dx\]
put \[e^x=t,e^x~dx=dt\]
\[I=\int\limits \frac{ dt }{ \sqrt{t^2-1} }\]
put t=sec z,dt=sec z tan z dz
and solve.

Answer 2

\[I=\int\limits \frac{ secz \tan z }{ \tan z }dz=\int\limits \sec z dz=\ln \left| \sec z+\tan z \right|+c=\ln \left| \sec z+\sqrt{\sec ^2 z-1} \right|+c\]

Answer 3

solve it

Answer 4

or you can also solve like this.
\[Put ~t=\frac{ 1 }{ z },dt=-\frac{ 1 }{ z^2 }dz\]
\[I=\int\limits \frac{ dt }{ \sqrt{t^2-1} }=\int\limits \frac{ \frac{ -1 }{ z^2 } dz}{ \frac{ \sqrt{1-z^2} }{ z } }\]
\[=-\int\limits \frac{ z~dz }{ z^2\sqrt{1-z^2} }\]
\[\sqrt{1-z^2}=w,1-z^2=w^2,z^2=1-w^2,2~z~dz=-2~w~dw,z~dz=-w~dw\]
\[I=\int\limits \frac{ w~dw }{ \left( 1-w^2 \right) w}=\int\limits \frac{ dw }{ \left( 1+w \right)\left( 1-w \right) }\]
\[=\int\limits \frac{ dw }{ 2\left( 1-w \right) }+\int\limits \frac{ dw }{ 2\left( 1+w \right) }=\frac{ -1 }{ 2 }\ln \left| 1- w \right|+\frac{ 1 }{ 2 }\ln \left| 1+w \right|\]
\[=\frac{ 1 }{ 2 }\ln \frac{ \left| 1+w \right| }{ \left| 1-w \right| }=\frac{ 1 }{ 2 }\ln \frac{ \left| 1+\sqrt{1-z^2}\right| } { \left| 1-\sqrt{1-z^2} \right|}\]
\[=\frac{ 1 }{ 2 }\ln \frac{ \left| 1+\sqrt{1-\frac{ 1 }{ t^2 }} \right| }{ 1-\sqrt{1-\frac{ 1 }{ t^2 }}}\]
\[=\frac{ 1 }{ 2 }\ln \frac{ \left| t+\sqrt{t^2-1} \right| }{ \left| t-\sqrt{t^2-1} \right| }\]
\[=\frac{ 1 }{ 2 }\ln \frac{ \left( t+\sqrt{t^2-1} \right)^2 }{ t^2-\left( t^2-1 \right) }=\ln \left( t+\sqrt{t^2-1} \right)\]
\[=\ln \left( e^x+\sqrt{e ^{2x}-1} \right)+c\]

Answer 5

@surjithayer - How do you take \[e ^{x} \] out of \[\sqrt{1-e ^{-2x}}\] ??

Answer 6

\[***e^{-x}\]

Answer 7

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