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Calculus demystified for Real People, Part 4 -- Formulae Rearranged
Integration By Parts
Since we started to study ourselves this crazy math..we wanted to find ways to simplify 'Heuristically' using easier Methods and our own Mnemonics instilled in us by our Tutor... That our Dad got us in Junior High. Lucky us right?
STRAIGHT TO our Subject
If we really need a practical formula for
We postulate a new arrangement as in the formula below. You can obtain this new formula by Integrating both sides of the equation
We've got to make sure that f can be differentiated to f′
and g be able to get integrated as well...
We should know that ∫(1/x2)dx = -1/x
and ∫x-1dx = ln/x/ +c
Consequently Inversely , we can asses that:
(ln x) ′= 1/x ( differentiation of natural logarithm for x)
Let's say we want to integrate this function....we know beforehand that we can break it down...
We rearrange it to our formula of 2 functions
∫(lnx)(x-2 )dx= lnx ∫x-2dx- ∫ f′ (∫g)dx
where f′ equals to 1/x
and g integrated (∫g) equals to -1/x
replacing in the formula for integration of two functions of the type ∫f.gdx=??
we get the easier to solve equation:
The answer is
You should know by now that
∫x^k or ∫xk = xk+1/k+1
(xk)′= kxk-1 (differentiation of a power)