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

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

Postby Integr88 » Wed 30 March, 2011 09:33

The question says integrate `int(x^2+x+1)/(sqrtx)dx` show all steps.
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Re: how to integrate `int(x^2+x+1)/(sqrtx)dx`

Postby barnamah » Wed 30 March, 2011 10:06

to integrate `int(x^2+x+1)/(sqrtx)dx` first because `sqrtx` is at the denominator of all terms, we can rewrite it as:

`int(x^2/sqrtx+x/sqrtx+1/sqrtx)dx`
Now it looks simpler to integrate so we can integrate each term:
`int(x^2/sqrtx)dx + int(x/sqrtx)dx+int(1/sqrtx)dx`


1-let's go now integrating it terms by term.
for first term `int(x^2/sqrtx)dx`

use u substitution
`u=sqrtx`
`du=1/2*1/sqrtxdx`
and because we need to have `1/sqrtx` in our equation we multiply both side by 2 and have:
`2du=1/sqrtxdx`

and we set `u=sqrtx` then `x=u^2` so we sub our values in our equation `int(x^2/sqrtx)dx`

`int(x^2/sqrtx)dx=int((u^2)^2*2)du=2int(u^4)du`
Now integrate it: `2*u^(4+1)/(4+1)=2*u^5/5=2/5u^5+C`

sub in the u we used above `2/5(sqrtx)^5+C`

2-Let's integrate `intx/sqrtxdx`

use u substitution
`u=sqrtx`
`du=1/2*1/sqrtxdx`
and because we need to have `1/sqrtx` in our equation we multiply both side by 2 and have:
`2du=1/sqrtxdx`

and we set `u=sqrtx` then `x=u^2` so we sub our values in our equation `int(x^2/sqrtx)dx`
`intu^2*2du=2intu^2du`

our equation is now `2intu^2du`
Integrate is `2*u^(2+1)/(2+1)=2u^3/3=2/3u^3+C`

sub in the u we used above `2/3(sqrtx)^3+C`

3-Let's integrate `int(1/sqrtx)dx`
we can write `int(1/sqrtx)dx=intx^(-1/2)dx`
`sqrtx=x^(1/2)` and `1/sqrtx=x^(-1/2)`


Integrating it, it looks `x^(-1/2+1)/(-1/2+1)=x^(1/2)/(1/2)=2*x^(1/2)=2sqrtx+C`

Now put it all together:


`2/5(sqrtx)^5+2/3(sqrtx)^3+2sqrtx+C`
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