## sjg13e one year ago Integration by Parts problem

1. sjg13e

$\int\limits_{}^{}\tan^{-1}xdx$

2. anonymous

try u = arctan x dv = dx

3. sjg13e

$\let u = \tan^{-1} and du = \frac{ -1 }{ x^2 +1 }dx$

4. sjg13e

then dv = dx and v = x

5. sjg13e

so, $\tan^{-1} \frac{ 1 }{ x } x - \int\limits_{}^{}xdx = \tan^{-1} \frac{ 1 }{ x } x - .5x^2$

6. anonymous

* du should be positve du 1/(x²+1)

7. sjg13e

why is du positive?

8. anonymous

that's the derivative of arctan x. $\frac{ d }{ dx }\tan^{-1} x=\frac{ 1 }{ 1+x^2 }$

9. sjg13e

yes but chain rule

10. sjg13e

we need to take the derivative of the inside function, which is 1/x

11. anonymous

no. that's an inverse function, not an exponent

12. sjg13e

$(1/x)\prime = -1/x^2$

13. sjg13e

i think it's treated as an exponent

14. sjg13e

1/x = x^-1 = -1/2 * x^2

15. sjg13e

ohh I'm sorry!! i forgot to properly write the initial problem. It's supposed to say $\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x }) dx$

16. sjg13e

i apologize. that's totally my fault

17. anonymous

oh ok

18. sjg13e

okay, ahah, so sorry about that! so besides that little mishap, does everything else look right?

19. anonymous

yeah it looks good

20. sjg13e

okay thank you!

21. anonymous

yw

22. sjg13e

Just for future reference. My previous answer was wrong, here's the correct, complete solution: $\int\limits_{}^{}\tan^{-1} (\frac{ 1 }{ x })dx = x \tan^{-1}(\frac{ 1 }{ x }) - \int\limits_{}^{}x*\frac{ 1 }{ x^2 +1 }dx$ at this point, you have to do u-subsition in the integral. so, let u = x^2 + 1, which will make .5du = xdx we can bring the - and the .5 out front to give: $x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ du }{ u }$ $=x \tan^{-1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\ln|u|$ $= x \tan^{-1} (\frac{ 1 }{ x }) + \ln |x^2 + 1| + C$