Question #a5322

2 Answers
Jan 18, 2018

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.

Jan 18, 2018

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