Question

How do you integrate `int (z^2+1)/sqrt(z)dz`?

Answer 1

I found: `2(z^(5/2)/5+sqrt(z))+c`

Explanation:

Let us set `z=t^2` so:
`dz=2tdt`
and substituting:
`=int(t^4+1)/cancel(t)(2cancel(t))dt=`
`2int(t^4+1)dt=2(t^5/5+t)+c=`
back to `z`:
`=2(z^(5/2)/5+sqrt(z))+c`

Answer 2

`2/5 z^(1/2) ( z^(4/2) +5) + C => 2/5 sqrt(z) (z^2+5)+C`

Explanation:

Separate and simplify the original expression into two parts then integrate each part. Sum the final integrations:

Write `(z^2+1)/(sqrt(z)) " "` as `" "(z^2)/(z^(1/2)) + 1/(z^(1/2))`

Simplifying the indices giving:

`z^(3/2) + z^(-1/2)`

Now integrate so we have:

`int(z^(3/2)).dz + int(z^(-1/2)).dz`

Giving:

`[ (z^(3/2 +2/2))/(5/2) + C_1 ] + [ (z^(-1/2 + 2/2))/(1/2) + C_2]`

Let `C_1 + C_2 = C_3` giving:

`2/5 x^(5/2) + 2z^(1/2) + C_3`

But `2/5 z^(1/2) times 5 = 2z^(1/2)`

Factoring out `2/5 z^(1/2)` gives:

`2/5 sqrt(z) (z^2+5)+C_3`