Question

How do you integrate `int (x^2+x+1)/sqrtxdx`?

Answer 1

`int (x^2+x+1)/sqrt(x)dx=(2/5x^3+2/3x^2+2x)/sqrt(x)`

Explanation:

As the integral is a linear operator, we can integrate term by term:

`int (x^2+x+1)/sqrt(x)dx= int x^2/sqrt(x)dx+int x/sqrt(x)dx+int 1/sqrt(x)dx = int x^(3/2)dx+intx^(1/2)dx +int x^(-1/2)dx = 2/5x^(5/2)+2/3x^(3/2)+2x^(1/2)= (2/5x^3+2/3x^2+2x)/sqrt(x)`

Answer 2

The answer is `=2(x^(5/2)/5+x^(3/2)/3+sqrtx)+C`

Explanation:

We need

`intx^ndx=x^(n+1)/(n+1)+C ( n!=-1)`

Therefore,

`int((x^2+x+1)dx)/sqrtx`

`=int(x^2/sqrtx+x/sqrtx+1/sqrtx)dx`

`=int(x^(3/2)+sqrtx+x^(-1/2))dx`

`=x^(5/2)/(5/2)+x^(3/2)/(3/2)+x^(1/2)/(1/2)+C`

`=2(x^(5/2)/5+x^(3/2)/3+x^(1/2))+C`