S(t√t+√t)dt/2

Answer 1

To solve the integral ∫S(t√t + √t) dt/2, we first simplify the integrand by factoring out a √t:

∫S(t√t + √t) dt/2
= ∫S(√t(t + 1)) dt/2
= ∫S√t(t + 1) dt/2

Next, we apply the u-substitution method by letting u = √t. Then, du/dt = 1/(2√t) and du = dt/(2√t). We can then express dt in terms of du as dt = 2√t du:

∫S√t(t + 1) dt/2
= ∫S(u^2(u^2 + 1)) 2√t du/2
= ∫S(2u^4 + u^2) du

Integrating term-by-term, we find:

= ∫S2u^4 du + ∫Su^2 du
= (2/5)u^5 + (1/3)u^3 + C
= (2/5)(√t)^5 + (1/3)(√t)^3 + C
= (2/5)t^{5/2} + (1/3)t^{3/2} + C

Therefore, the result of the integral ∫S(t√t + √t) dt/2 is (2/5)t^{5/2} + (1/3)t^{3/2} + C, where C is the constant of integration.