Question

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

Answer 1

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`