## AbdulRehman Group Title what will be the integration of 1/(1+x²)½ along x? 3 years ago 3 years ago

1. Atooty-S

$\int\limits 1/(1+x ^{2})^{1/2}$ it will be $\ln (1 + x ^{2})^{1/2}$

2. Atooty-S

you know that when the numerator is the derivative of the dominator then the integration is ln "dominator" :)

3. uzma

but the derivative of denominator should be there in the numerator

4. uzma

$x= \tan \theta$

5. Atooty-S

ok , the derivative of (x^2 )^1/2 is 1/2 (2x) so 2 cancel the other 2 :)

6. uzma

and for natural log, exponent should be 1

7. uzma

$\int\limits 1/x= \ln x$

8. uzma

if the exponent is other than 1, we use power rule

9. AbdulRehman

yeah but we can't multiply it with 2x so ln(1+x²)½ is not the right answer

10. uzma

$d x= \sec^2 \theta d \theta$

11. Atooty-S

i think its correct .. anyways good luck ..

12. uzma

$\int\limits \sec^2 \theta/(\sqrt{1+\tan^2 \theta} )$

13. uzma

$\int\limits \sec^2\theta/\sec \theta d \theta$

14. uzma

$\int\limits \sec \theta d \theta$

15. uzma

$\ln (\sec \theta + \tan \theta) +C$

16. uzma

by backward substitution $\tan \theta =x$

17. AbdulRehman

yeah that will give us integration of secθ that is ln(secθ+tanθ) and by substituting and we will get probably get ln(sec(tan^-x)+x)

18. uzma

$\sec \theta =\sqrt{1+x^2}$

19. AbdulRehman

thanks for the help i have some questions on multiple integral usage in finding area and volume. Anyone please help me

20. uzma

post it as a new Q,some one would surly look at it :)

21. AbdulRehman

Uzma are you from Pakistan?

22. Atooty-S

ahaa ! ok waiting ur new Qs :)

23. uzma

yes :)

24. AbdulRehman

25. uzma

what would u do with that :)

26. AbdulRehman

nothing just to guess the level of your understanding mathematics

27. uzma

mathematics is as vast, that one can hardly grasp even one area