# Evaluate `int` (t)^1/30 dt Answer=

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### 2 Answers

We can use the formula:

`intx^ndx = x^(n+1)/(n+1) + c`

Here, n = 1/30.

`intt^(1/30)dt = t^(1/30 + 1)/(1/30 + 1) + c`

The lcd here is 30. So, we will multiply the 1 by 30/30 to make

the denominators equal.

`t^(1/30 + 30/30)/(1/30 + 30/30) + c = t^(31/30)/(31/30) + c`

Flip the bottom and proceed to multiplication.

`t^(31/30) *(30/31) = (30t^(31/30))/31 + c`

`int t^(1/30) dt=`

since `int( x^k) dx= 1/(k+1) x^(k+1)+c` we get:

`int t^(1/30) dt= 1/(1/30+1) x^(1/30+1)=30/31 x^(31/30)+c` `=30/31 x root(30)(x) +c`