How do you find the integral of #20+ (4s^4)/sqrts ds#?

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Answer 1 · 2016-06-03T01:50:38

#20s+8/9s^(9/2)+C#

We have

#int(20+(4s^4)/sqrts)ds#

Split this up since integrals can be separated through addition:

#=int20ds+int(4s^4)/sqrtsds#

The second integrand can be simplified as follows:

#(4s^4)/sqrts=(4s^4)/s^(1/2)=4s^(4-1/2)=4s^(7/2)#

So we have the integral

#int20ds+4ints^(7/2)ds#

The first integral is just #20s+C#, and find the next integral using the rule #ints^nds=s^(n+1)/(n+1)+C#.

#=20s+4(s^(7/2+1)/(7/2+1))+C#

#=20s+4(2/9)s^(9/2)+C#

#=20s+8/9s^(9/2)+C#