2 Integrations by subs. qu's (1 Viewer)

[currysauce]

Hey, if anyone could do these, thanks.

1. By using the subsitution u² = x + 2, find

integral [ x-2 / root(x+2) ] dx



2. Use the substitution u = x + 4 to find

integral [ 5x root(x+4) ] dx

 

[withoutaface]

/
| [(x-2)/&sqrt;(x+2)]dx
/

/
| [(u²-4)/u]du
/

/
|1 - 4/u du
/

u -4ln(u)+C

&sqrt;(x+2) -2ln(x+2)+C

Currently working on 2nd one

/
| 5(u-4) sqrt(u)du
/
/
|5u^3/2-20u^1/2
/

2u^5/2 - 40u^3/2/3+C

2(x+4)^5/2 - 40(x+4)^3/2/3+C

I'm tired and there's probably an error in there, but you get the idea...

 

[currysauce]

your 2nd one checks out to be correct, thanks.

Also the first one has no, ln ,i n it.

the answer is exact (has roots and all).

but thanks for da 2nd qu!.

 

[QuaCk]

1. By using the subsitution u² = x + 2, find

integral [ x-2 / root(x+2) ] dx


u^2 = x + 2 => x = u^2 - 2
2u.du = dx


integral [ x-2 / root(x+2) ] dx
integral [ (u^2 - 4) / u ] 2u.du
integral 2 [ u^2 - 4 ] du
2 [ u^3/3 - 4u ] + C

and sub it in ...

if i'm not mistaken ....

 

[illin]

Quack is right, not withoutaface
withoutaface: u forgot about du/dx, cause du (does not =) dx
but u did get no2 right