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Is there such thing as the integral of u/du?
I have heard about the integral of u du... but not u over du...
The original exercise is the integral of tan^-1(k) / 1 + k^2....
So, I am 100% certain that u = tan^-1 (k), and that therefore du = 1 + k^2... but that would imply that du is in the denominator... what do I do from here?
1 Answer
- PrashantLv 68 years agoFavorite Answer
There is some problem in your understanding.
Firstly,
the question must be integral of tan^-1(k) / 1 + k^2 * dk , not integral of tan^-1(k) / 1 + k^2
(The 'dk' is important.)
Next,
u = tan^-1 (k)
So u is a function of k.
Differentiate u w.r.t. k:
du / dk = 1 / 1+k^2.
or
du = dk / (1+k^2)
Now substitute in the integral:
Integral of u du
Hope this helps.
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