∫ [ln(√t) / t] dx
let √t = u
t= u² → dx = 2u du
substitute in the integral
∫ [ln(√t) / t] dx = ∫ (ln u / u²) 2u du = ∫ (ln u / u²) 2u du = 2 ∫ (ln u / u) du
let ln u = x → d (ln u) = dx→ (1/u)du = dx
substituting again
2 ∫ (ln u / u) du = 2 ∫ x dx= 2 x²/ 2 = x² + c which,
substituting ln² u + c
as of the first
substitution ln²(√t) + c
it concludes that
∫ [ln(√t) / t] dx = ln²(√t) + c
hope it helps
