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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

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