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 $ints^nds=s^(n+1)/(n+1)+C$ .

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

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

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

How do you find the integral of 20+ (4s^4)/sqrts ds20+ (4s^4)/sqrts ds?