Integral : (x^2)/((x^2+1)^2)

  • Thread starter Alexx1
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  • #1
Alexx1
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The answer of the integral of (x^2)/((x^2+1)^2) is (1/2)(arctan(x)-(x/x^2+1))

In class, we've seen the steps to solve this integral, but I don't understand certain steps..
Can someone explain me how to solve this integral, step by step?
 

Answers and Replies

  • #2
uart
Science Advisor
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If you can post the steps and point out those that you didn't understand then I'm sure someone can help you.

BTW. The easiest way to do that one is "integration by parts". Have you learnt this technique yet?
 
  • #3
Alexx1
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If you can post the steps and point out those that you didn't understand then I'm sure someone can help you.

BTW. The easiest way to do that one is "integration by parts". Have you learnt this technique yet?

Sure, no problem, here are the steps:

Integral((x^2)/((x^2+1)^2)dx)
= (1/2)*Integral(x d(1/(x^2+1))
= (1/2)*(x/(x^2+1))-(1/2)*Integral(1/(x^2+1)dx)
= (1/2)*(x/(x^2+1))-(arctan(x))/2

Last step is the answer

(The answer I said earlier was wrong, this is the correct answer:(1/2)*(x/(x^2+1))-(arctan(x))/2)

Thank you
 
  • #4
Bohrok
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[tex]\int\frac{x^2}{(x^2 + 1)^2} = \int x \frac{x}{(x^2 + 1)^2}[/tex]

Using ∫u v' = uv - ∫v u',
let u = x and v' = x/(x2 + 1)2
 
  • #5
Dunebug7
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It's basically separating it into parts ie.


[itex]\int \frac{x^2}{(x^2+1)^2}\rightarrow \int \frac{x}{1}.\frac{x}{(x^2+1)^2}\equiv x(x. \sin(\arctan(x)))[/itex]

as

[itex]\frac{x}{1}=\frac{1}{2}x^2[/itex]

and [itex]x\frac{x}{(1+x)^2}=x.\sin(\arctan(x))[/itex]

By the trig identity.

Thus the answer is:

[itex]\int\frac{x^2}{(x^2+1)^2}=-\frac{1}{2}.\frac{x}{(x^2+1)}+\frac{1}{2}\arctan(x)+C[/itex]

Don't forget the constant of integration, it's a silly way to loose marks. :smile:
 
Last edited:
  • #6
Alexx1
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Thank you both!
 
  • #7
Dunebug7
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Thank you both!

np Bhorok's answer is more elegant and easier, but I thought you might need a long winded explanation and there's often more than one way to swing a cat I guess. Hope it helped. :smile:
 
  • #8
fourier jr
757
13
The answer of the integral of (x^2)/((x^2+1)^2) is (1/2)(arctan(x)-(x/x^2+1))

In class, we've seen the steps to solve this integral, but I don't understand certain steps..
Can someone explain me how to solve this integral, step by step?

since you have the answer, take its derivative & work backwards. that's how to figure it out. just don't show anyone your rough work :tongue2:
 
  • #9
uart
Science Advisor
2,797
21
Sure, no problem, here are the steps:

Integral((x^2)/((x^2+1)^2)dx)
= (1/2)*Integral(x d(-1/(x^2+1))
= (-1/2)*(x/(x^2+1)) -(1/2)*Integral(1/(x^2+1)dx)
= -(1/2)*(x/(x^2+1))+(arctan(x))/2

Last step is the answer

(The answer I said earlier was wrong, this is the correct answer:(1/2)*(x/(x^2+1))-(arctan(x))/2)

Thank you

No the original answer was correct, you dropped a minus sign in the first line of this derivation.
 

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