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How do you find the indefinite integral of #int 4/sqrt(5t)dt#?

1 Answer

Apr 13, 2017

THe answer is #=(8sqrtt)/sqrt5+C#

Explanation:

We need

#intx^ndx=x^(n+1)/(n+1) +C#

#int1/sqrtxdx=intx^(-1/2)dx=x^(1/2)/(1/2)=2sqrtx#

So,

#int(4dt)/sqrt(5t)=4/sqrt5*2sqrtt+C#

#=(8sqrtt)/sqrt5+C#

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