have i forgotten? (1 Viewer)

[S.M.R.]

I was wondering how to integrate f(x) = (3x+1) / (x+1)^(1/2) (sorry idk where the equation editor is...). I have forgotten integration, such are the perils of the hsc lol. I was thinking you cant do the old wack-the-derivative-on-the-top-and-slap-a-ln-on-the-side method, is partial functions the way to go? thx in advance

 

[Timothy.Siu]

I was wondering how to integrate f(x) = (3x+1) / (x+1)^(1/2) (sorry idk where the equation editor is...). I have forgotten integration, such are the perils of the hsc lol. I was thinking you cant do the old wack-the-derivative-on-the-top-and-slap-a-ln-on-the-side method, is partial functions the way to go? thx in advance

uhh

S (3x+1)/(x+1)^(1/2) dx

=3 S (x+1-2/3)/(x+1)^1/2 dx

=3 S (x+1)^(1/2)dx- 2/3 S 1/(x+1)^1/2 dx

S is the integration sign lol, then ur done

 

[Drongoski]

 

[gurmies]

 

[S.M.R.]

it was just substitution? methinks i need to do some sleep, or hamlet (nahh)

 

[gurmies]

Well, substitution is really the dropkick method in my opinion, for questions where it's not necessary (as demonstrated in the above posts). Having said that, I often make mistakes with that method, so sometimes a substitution is okay...