Question

Question #a5322

Answer 1

`8(ln(sqrt(1+e^(2x) ) - 1) - x) + C`

Explanation:

Let `u=sqrt(1+e^(2x))` i.e.

`du = (e^(2x))/(u)dx`
`dx = 1/e^(2x)udu = (udu)/(u^2 - 1)`

Therefore, the integral becomes

`int 8/sqrt(1 + e^(2x))dx = int 8/u (udu)/(u^2-1) = int (8du)/(u^2 - 1)`

We can use partial fraction decomposition to separate this out, yielding
B = 4, A = -4
`int (8du)/(u^2 - 1) = int (4/(u-1) - 4/(u+1))du = 4ln((u-1)/(u+1)) + C`

Miraculously, this is the same as your above statement!

`4ln((u-1)/(u+1)) = 8[1/2ln(u-1) - 1/2 ln(u+1)]`
`= 8[ln(u-1) - 1/2ln(u^2-1)] = 8[ln(u-1) - x]`
`= 8[ln(sqrt(1+e^(2x)) - 1) - x]`

where the constant of integration has been left out for convenience.

Answer 2

The answer is `=8ln(sqrt(e^(2x)+1)-1)-8x+C`

Explanation:

Perform the substitution

`u=1+e^(2x)`, `=>`, `du=2e^(2x)dx`

`e^(2x)=u-1`

Therefore,

`int(8dx)/(sqrt(1+e^(2x)))=8int(du)/(2(u-1)sqrtu)`

`=4int(du)/(sqrtu(u-1))`

Perform the substitution `v=sqrtu`, `=>`, `dv=(du)/(2sqrtu)`

`int(8dx)/(sqrt(1+e^(2x)))=8int(dv)/(v^2-1)`

Perform the decomposition into partial fractions

`1/(v^2-1)=A/(v+1)+B/(v-1)=(A(v-1)+B(v+1))/(v^2-1)`

`1=A(v-1)+B(v+1)`

Let `v=-1`, `1=-2A`, `=>`, `A=-1/2`

Let `v=1`, `1=2B`, `=>`, `B=1/2`

So,

`int(8dx)/(sqrt(1+e^(2x)))=8int-1/2(dv)/(v+1)+8int1/2(dv)/(v-1)`

`=-4ln(v+1)+4ln(v-1)`

`=-4ln(sqrtu+1)+4ln(sqrtu-1)`

`=-4ln(sqrt(1+e^(2x))+1) + 4ln (sqrt(1+e^(2x))-1))`

`=4ln(((sqrt(e^(2x)+1))-1)/((sqrt(e^(2x)+1))+1))`

`=4ln((((sqrt((e^(2x)+1))-1)(sqrt(e^(2x)+1))-1))/(((sqrt((e^(2x)+1)))+1)(sqrt(e^(2x)+1))-1))`

`=4ln(((sqrt(e^(2x)+1)-1))^2/(e^(2x)))`

`=8ln(sqrt(e^(2x)+1)-1)-4ln(e^(2x))`

`=8ln(sqrt(e^(2x)+1)-1)-8x+C`