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what will be the integration of 1/(1+x²)½ along x?

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\[\int\limits 1/(1+x ^{2})^{1/2}\] it will be \[\ln (1 + x ^{2})^{1/2}\]
you know that when the numerator is the derivative of the dominator then the integration is ln "dominator" :)
but the derivative of denominator should be there in the numerator

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Other answers:

\[x= \tan \theta\]
ok , the derivative of (x^2 )^1/2 is 1/2 (2x) so 2 cancel the other 2 :)
and for natural log, exponent should be 1
\[\int\limits 1/x= \ln x\]
if the exponent is other than 1, we use power rule
yeah but we can't multiply it with 2x so ln(1+x²)½ is not the right answer
\[d x= \sec^2 \theta d \theta\]
i think its correct .. anyways good luck ..
\[\int\limits \sec^2 \theta/(\sqrt{1+\tan^2 \theta} )\]
\[\int\limits \sec^2\theta/\sec \theta d \theta\]
\[\int\limits \sec \theta d \theta\]
\[\ln (\sec \theta + \tan \theta) +C\]
by backward substitution \[\tan \theta =x\]
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)
\[\sec \theta =\sqrt{1+x^2}\]
thanks for the help i have some questions on multiple integral usage in finding area and volume. Anyone please help me
post it as a new Q,some one would surly look at it :)
Uzma are you from Pakistan?
ahaa ! ok waiting ur new Qs :)
yes :)
Can you please tell me your qualification and city
what would u do with that :)
nothing just to guess the level of your understanding mathematics
mathematics is as vast, that one can hardly grasp even one area

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