Question

How do you integrate `(4x^2 + 6x^(3/2) + 13x + 5)/sqrtx`?

Answer 1

`8/5x^(5/2)+3x^2+26/3x^(3/2)+10x^(1/2)+C`

Explanation:

Rearrange the algebra to make it all one continuous line, which is far easier to integrate than having a fraction, knowing that `sqrtx = x^(1/2)`

`(4x^2+6x^(3/2)+13x+5)/x^(1/2)=4x^(2-1/2)+6x^(3/2-1/2)+13x^(1-1/2)+5x^(0-1/2)`
`=4x^(3/2)+6x+13x^(1/2)+5x^(-1/2)`

Now this is far easier to integrate. Increase each power by one and divide by the new power, doing each phrase individually and putting them back together at the end. At this point we can leave out the constant of integration, but make sure to replace it at the end.

`int4x^(3/2)dx=(4x^(5/2))/(5/2) = 8/5x^(5/2)`
`int6xdx=3x^2`
`int13x^(1/2)dx=26/3x^(3/2)`
`int5x^(-1/2)dx=10x^(1/2)`

Now put these back together, and add in the constant of integration,

`8/5x^(5/2)+3x^2+26/3x^(3/2)+10x^(1/2)+C`

which is the final answer.

You can check by differentiating this function and checking it is the same as the original question.