Integration Help

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Homework Statement


[tex]\int\frac{2x+1+\sqrt{x} dx}{(x+1)(1+\sqrt{x}}[/tex]

Homework Equations





The Attempt at a Solution


I've tried substitution using u=x+1, u=1+sqrt(x) but every time I try I get 3 simple integrals + [tex]\int\frac{dx}{(x+1)(1+\sqrt{x})}[/tex]

Is there any way to avoid partial fractions? I really hate doing those...also it seems (x+1)(1+sqrt(x)) will give a wierd expansion.
 

Answers and Replies

  • #2
Dick
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You may hate partial fractions, but they speak quite well of you. It's an easy way to exploit that factored denominator. And substituting u^2=x won't hurt either. It's really a polynomial problem in disguise.
 
  • #3
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hm..thanks again lol I tried u^2=x but gave up too soon on it before...
Just got like..6 really simple integrals now.

ok just got into this fraction doing the separation of the top after i did the substitution:

[tex]\frac{1}{(u^2+1)(1+u)}=\frac{A}{u^2+1}+\frac{B}{1+u}[/tex]

so:

[tex]1=A(1+u)+B(u^2+1)[/tex]
plugging in u=-1 I get
1=2B so B=.5

then to get A I tried u=1
1=2A+.5(2)
1=2A+1
0=2A
0=A

ok but if I try u=3...
1=4A+.5(10)
1=4A+5
-4=4A
-1=A

If I try u=5 I get A=-2....

I don't get why this is going on but also how can I solve this partial fraction? ><
 
Last edited:
  • #4
Dick
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If you have a quadratic factor like (u^2+1) you should allow the numerator to be a linear function like D*u+E. Otherwise you won't have enough variables to give you a solution. You knew that, didn't you? You've just let your hatred of partial fractions overwhelm you.
 
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  • #5
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ah...forgot that >< and I just looked over my calc 2 notes too.....

ok so:
[tex]1=(Au+B)(1+u) + C(u^2+1)[/tex]
[tex]1=Au(1+u)+B(1+u)+C(u^2+1)[/tex]
u=-1
1=2C
C=.5

u=0
1=B+.5
B=.5

u=1
1=2A+1+1
1=2A+2
-1=2A
A=-.5

so:
[tex]\frac{1}{(u^2+1)(1+u)}=\frac{-.5u+.5}{u^2+1}+\frac{.5}{1+u}[/tex]

ok now that looks right....I just really try to avoid partial fractions unless I MUST go that way, same thing with trig substitutions or u=hyperbolic function of x
 
  • #6
Dick
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It's a good idea to avoid hard things. But sometimes you HAVE to. And that's correct.
 
  • #7
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i start learning partial fractions next week

eek! be prepared to help me out bob! lol
 
  • #8
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Thanks.

hehe rocophysics just remember to not split up the RHS even though some teachers ask you to expand. just leave the (u^2+c) (u+c) etc... in that form and then try to get something to cancel that u^2 or u to get one of those unkown variables.

This one wasn't that bad but some you get like 3-4 partial fractions with like 5-8 uknown variables and you spend a while getting all of that..but that's rare :s
 
  • #9
Dick
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i start learning partial fractions next week

eek! be prepared to help me out bob! lol

If you pay attention, maybe no one will have to help you. And you can use that spare time to help others.
 
  • #10
Dick
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Thanks.

hehe rocophysics just remember to not split up the RHS even though some teachers ask you to expand. just leave the (u^2+c) (u+c) etc... in that form and then try to get something to cancel that u^2 or u to get one of those unkown variables.

This one wasn't that bad but some you get like 3-4 partial fractions with like 5-8 uknown variables and you spend a while getting all of that..but that's rare :s

If it gets that extreme, just use a machine. Like maxima or mathematica. They are really quite good at such stuff. But you have to understand how they did it. Otherwise you can't check them. And they do make mistakes.
 
  • #11
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If you pay attention, maybe no one will have to help you. And you can use that spare time to help others.
i didn't mean it literally. but i study hard, and this forum does motivate me to work hard and ahead so i can help others.
 
  • #12
Dick
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i didn't mean it literally. but i study hard, and this forum does motivate me to work hard and ahead so i can help others.

Sorry, that did come out sounding harsh, didn't it? Partial fractions aren't that hard, but people never seem to 'get' them. So when you've 'got' them you'll find plenty of people to help.
 
  • #13
learningphysics
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u^2 =x
2udu = dx

[tex]\int\frac{dx}{(x+1)(1+\sqrt{x})}[/tex]

Did you substitute 2udu into the numerator?
 
  • #14
VietDao29
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u^2 =x
2udu = dx

[tex]\int\frac{dx}{(x+1)(1+\sqrt{x})}[/tex]

Did you substitute 2udu into the numerator?

Well, it isn't necessary, he's trying to Partial Fraction [tex]\frac{1}{(x + 1) (1 + \sqrt{x})}[/tex], after that, he can change u back to x, and work from there.
 
  • #15
learningphysics
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Well, it isn't necessary, he's trying to Partial Fraction [tex]\frac{1}{(x + 1) (1 + \sqrt{x})}[/tex], after that, he can change u back to x, and work from there.

Oh, that's right. My bad. Sorry.
 

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