Find the most general antiderivative of the function : (1 + t + t^2) / t^1/2

Question

Find the most general antiderivative of the function.

 

(1 + t + t²) / t^1/2

 

 

ANSWER: G(t)  =  2t^1/2 + 2/3t^2/3 + 2/5 t^5/2 + C

#15, p348

Answer 1

First, we divide each term in the numerator by the denominator.

 

∫( t^-1/2 + t^1/2 + t^3/2) dt

 

 

Now we can take the anti-derivative term by term.

 

∫t^-1/2 dt  + ∫t^1/2 dt  + ∫t^3/2 dt =

 

 

t^1/2 / (1/2) + t^3/2 / (3/2) + t^5/2 / (5/2) =

 

2t^1/2 + (2/3)t^3/2 + (2/5)t^5/2 + C

Answer 2

You can simply divide t^1/2 into each of the terms and then apply the power rule to each term (larger font so you can read the exponents more clearly):

 

∫ (1 + t + t²)dt / t^1/2 = ∫(t^-1/2 + t^1/2 + t^3/2)dt =

=t^1/2 / (1/2) + t^3/2 / (3/2) + t^5/2 / (5/2) + C =

=2t^1/2 + 2/3t^3/2+2/5t^5/2 + C