What is the antiderivative of #(t - 9t^2)/sqrt(t) dt#?

I got #(2/15)[t^(3/2)](5 - 27t) + C#. Is this correct?

2 Answers
Feb 7, 2018

#int (t-9t^2)/sqrtt dt=2/3t^(3/2)-18/5t^(5/2)+"c"#

Explanation:

#int (t-9t^2)/sqrtt dt = intt/sqrtt dt -9intt^2/sqrtt dt = intt^(1/2)dt -9int t^(3/2)dt#

We now use the power rule for integration:

#int x^n dx = 1/(n+1)x^(n+1)+"c"#

#therefore intt^(1/2)dt -9int t^(3/2)dt= 2/3t^(3/2)-18/5t^(5/2)+"c"#

And just to check, differentiating gives

#t^(1/2) -9t^(3/2) =(t-9t^2)/sqrtt#

Feb 7, 2018

# \ #

# \ 2/15 t^{3/2} ( 5 - 27 t ) + C. #

# \mbox{You Got It Exactly Right !! Bravo !!!} #

Explanation:

# \ #

# \mbox{This antiderivative allows a nice, quick rewrite of the integrand,} \ \ \mbox{which will then allow a thankfully smooth integration.} \ \ \mbox{Simplification later may be only slightly less smooth.} #

# \mbox{We have:} #

# \int \ \ { t - 9 t^2 } / \sqrt{t} dt \quad = \ \int \ \ ( t / \sqrt{t} - { 9 t ^2 } / \sqrt{t} ) dt #

# \qquad \qquad qquad \qquad qquad \qquad = \ \int \ \ ( t^{1/2} - 9 t^{3/2} ) dt #

# \qquad \qquad qquad \qquad qquad \qquad = \ 2/3 t^{3/2} - 9 ( 2/5 t^{5/2} ) + C. #

# \mbox{To simplify this now, factor out the numerical fractions using} \ \ \mbox{the LCM of their denominators and the GCD of their} \ \ \mbox{numerators, and factor out the common variables using their} \ \ \mbox{lowest common power: #

# \int \ \ { t - 9 t^2 } / \sqrt{t} dt \quad = 2/3 t^{3/2} - 9 ( 2/5 t^{5/2} ) + C #

# \qquad \qquad qquad \qquad qquad \qquad = \ 2/15 t^{3/2} (5) - \ 2/15 t^{3/2} ( 9 \cdot 3 t) + C. #

# \qquad \qquad qquad \qquad qquad \qquad = \ 2/15 t^{3/2} ( 5 - 27 t ) + C. #

# \ #

# \mbox{You Got It Exactly Right !! Bravo !!!} #