Question

Integrate this expression

Answer 1

the picture gives the answer

Answer 2

Answer:

Step-by-step explanation:

∫e^2x-e^-2x  dx / e^2x+e^-2x

dividing numerator and denominator with e^2x

∫1-e^-4x dx /1+e^-4x

∫dx/1+e^-4x + ∫-e^-4x dx /1+e^-4x

let y = 1+e^-4x

y-1=e^-4x

log y-1 =-4x

-1/4 log y-1 =x

∫dx/y

∫d(-1/4 log y-1) / y

∫-1/4 dy /y-1/y

-1/4∫dy/y-1*y

-1/4∫(1/y-1 - 1/y ) dy

-1/4{log y-1 - logy}

-1/4{log y-1/y}

-1/4{log e^-4x / 1+e^-4x}

∫dx/1+e^-4x = -1/4{log e^-4x / 1+e^-4x}

∫-e^-4x dx/1+e^-4x

1/4∫-4e^-4x dx /1+e^-4x

1/4log{1+e^-4x}

therefore ∫1-e^-4x/1+e^-4x

= -1/4{log e^-4x / 1+e^-4x}+1/4log{1+e^-4x}