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Integration by Parts

  • Thread starter dimpledur
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  • #1
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Homework Statement



[tex] \intx^2tan^{-1}xdx [/tex]



The Attempt at a Solution


[tex] \int{x^2tan^{-1}xdx} [/tex]



[tex]\int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-{\frac{1}{3}}\int \frac {x^3}{1+x^2}dx[/tex]



[tex]let {}u=1+x^2, \frac{du}{2}=xdx[/tex]



[tex]\frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int (1-1/u)[/tex]



[tex]\frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)[/tex]



In the answer, there is no -1/6. Help?

I'm not looking for other alternatives. I have the solution for this question, however, I want to know why the 1/6 dissapears
 
Last edited:

Answers and Replies

  • #2
194
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I think that's it.... I can't see the integral signs on this site..... Although I can on every other site that uses latex.
 
Last edited:
  • #3
4
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We know that arctan x is differentiable
u = arctan x
du/dx = 1/1+x^2

x^2dx = dv so v = x^3/3

now int udv = uv - int vdu/dx dx
= x^3(arctan x) - int (x^3/3(1+x^2) dx)

let 1+ x^ 2 = t
x^2 = (t-1)

2xdx = - dt
so we get x^3/1(1+x^2) dx
= x^2xdx/(1+x^2)/3
now you can proceed as
= (t-1) (-dt/6)/t
= -1/6(dt- dt/t)
integrating we get -1/6 t + 1/6ln(t) as t >0

substitute the value and get the result
 
  • #4
194
0
Nevermind. My answer was right. I didnt realize that terms without a variable were assumed to be with the constant C.
 
  • #5
33,521
5,199

Homework Statement



[tex] \intx^2tan^{-1}xdx [/tex]
Put a space between \int and what follows it, like so.
[tex] \int x^2tan^{-1}xdx [/tex]

The Attempt at a Solution


[tex] \int{x^2tan^{-1}xdx} [/tex]



[tex]\int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-{\frac{1}{3}}\int \frac {x^3}{1+x^2}dx[/tex]



[tex]let {}u=1+x^2, \frac{du}{2}=xdx[/tex]



[tex]\frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int (1-1/u)[/tex]



[tex]\frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)[/tex]



In the answer, there is no -1/6. Help?

I'm not looking for other alternatives. I have the solution for this question, however, I want to know why the 1/6 dissapears
 

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