Integration by Parts: Solving \int{x^2tan^{-1}xdx}

[dimpledur]

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

 

[dimpledur]

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.

 

[!^--_--^!]

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

 

[dimpledur]

Nevermind. My answer was right. I didnt realize that terms without a variable were assumed to be with the constant C.

 

[Mark44]

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