Integration by parts tanx help

For the first problem you need to use integration by partial fractions to solve it.

You were on the right track in the second problem changing (tanx)^2 to (secx)^2 - 1 now you just have to integrate each seperatly which you did right except for the integral of (secx)^2 is just tanx

 

Quicker solution to the first problem:

Use variable substitution: [tex] u = x+2[/tex]
You'll then get:
[tex] \int \frac{something}{u^2}du[/tex]
which you can then separate into powers of u and integrate.

Your integration in the second problem is not quite correct.
Recall that [tex]\frac{d}{dx} \tan (x) = \sec^2(x)[/tex] not tan squared!!!

 

You got [tex]
\int\frac{x}{(x+2)^2}dx[/tex]

use the substitution u=x+2, so x=u-2 and dx=du and take it from there

 

You must finish the substitution.... rewrite x in terms of u.

[edit: And of course when you're done integrating, rewrite your answer back in terms of x.]

 

Right! (But don't forget the +C!!!) ;-)

 

Hi carbz!

A partial fractions method (which avoids having to substitue and then substitute back again ) is:

[tex]\int\frac{x}{x^{2}+4x+4}dx[/tex]

[tex]=\,\int\frac{x+2}{x^{2}+4x+4}dx
\,-\,\int\frac{2}{x^{2}+4x+4}dx[/tex]

 

You've not solved it yet. You'll still need to do a substitution to integrate the partial fractions. Now you must do it twice. Also since its a linear substitution you can do it in your head once you're practiced.

 

his thing = (x+2)/(x+2)^2 - 2/(x+2)^2
so, no *twice* substitution

 

No … jambaugh is right … you and I can do it in our heads, but without that knack it does require a substitution!

 

actually couldnt you do it with partial fractions and integrating once?

you can break up x/(x^2+4x+4) into x/(x+2)^2

therefore by partial fractions you could break it into [x/(x+2)^2]= A/(x+2) + Bx+C/[(x+2)^2]

from there you can solve for A, B, and C and plug that back into your original integral

not sure if thats what you originally ment but idk.

 

Hi tron_2.0!

Yes … and B will be 0 …

… but that's exactly what we have done!

 

ahh my mistake, i should have read the replies more closely =[

 

Nice one carbz

 

OK I take back the "twice", as it is the same substitution and you can write the two summed integrals as a single one.

In effect it is the same two steps in different orders as when I say "you can then separate into powers of u and integrate" this is a trivial case of expansion into partial fractions. Ultimately the preference is one of which you can do more easily the first substitution or the first expansion by partial fractions.