Indefinite integration using substitution

∫ (e^√t) / √t(e^√t+1) dt

u = e^(sqrt(t)) + 1
du = e^sqrt(t) * dt / (2 * sqrt(t))

e^(sqrt(t)) * dt / (sqrt(t) * (e^sqrt(t) + 1)) =>
2 * du / u

Integrate

2 * ln|u| + C =>
2 * ln|1 + e^sqrt(t)| + C

u = e^√t + 1
du = (e^√t)/(2√t) dt

∫ (e^√t) / (√t (e^√t+1)) dt = 2 ∫ 1/(e^√t+1)) (e^√t)/(2√t) dt
. . . . . . . . . . . . . . . . . . = 2 ∫ 1/u du
. . . . . . . . . . . . . . . . . . = 2 ln(u) + C
. . . . . . . . . . . . . . . . . . = 2 ln(e^√t + 1) + C

∫e^(√t)/(√t)(e^(√t) + 1) dt

z = e^(√t) => 2 dz = e^(√t)/t dt

2∫dz/(z + 1) = 2ln(e^(√t) + 1) + C

w = e^(√t) + 1...use it to generate an ln function