Question

Integrate √e^x -1
Integration of given equation

Answer 1

Answer:

Step-by-step explanation:

1/2√e^x-x+c

Answer 2

Answer:

Step-by-step explanation:

∫ √ e^x-1 dx

let e^x-1 =t

e^x=t+1

x=log t+1

∫ √t d(log t+1)

∫√t  dt / t+1

∫t dt / root(t) (t+1

∫t+1-1 dt / root(t) (t+1)

∫t+1 dt / root (t) (t+1) -∫dt/root(t)*(t+1)

∫dt/√t -∫dt/root(t)*t+1

∫t^-1/2 dt -∫dt/root(t)*t+1

2√t -∫dt/root(t)*t+1

∫dt/root(t)*t+1

let u= root t

du=dx/2roott

2roott du =dx

∫2√t du /√t (u^2+1)

2arctanx

=2√t -∫dt/root(t)*t+1

=2√t-2arctanu

=2√e^x-1 - 2arctan (√e^x-1)+c