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give MEDALS as much as possible :P just joking

that will be beneficial for me .. thanks lgba

welcome

Chain Rule , ........ etc ??

POINT NOTED.

yes.. i forgot chain rule..
\[\frac{d}{dx} (u) = u'du\]

it goes like that right?

Note in everything i wrote, u and v stand for FUNCTIONS of X

however, like i said...the things i wrote above are the most essential

GooD work.

oh. ill keep those points in mind. thanks!

Thanks from me too... they are really benifitiel ... i'll keep it in my mind... thanks again

yeah...that's probably the least possible lol

\[f'(x) = {\lim_{h \to 0} \;\;{f(x + h) - f(x) \over h}} \]

\[f \prime(x)=\lim_{x \rightarrow x _{0}}\frac{ f(x)-f(x_{0}) }{ x-x _{0} }\]

@ParthKohli, pop quiz!
Evaluate\[\int\frac{2x}{x^2+6}\,dx.\]:)

O Lord, please help me figuring out the technique I should use here. Trig substitution?

No... not trig substitution

jst substitute x^2 = u

Yeah, right!\[du = 2xdx\]

It's ln(x^2+6). You can use the f'(x)/f(x)case.

Woohoo!\[\int {1 \over u + 6}du\]

x^2+6=u will be more beneficial.

\[\int (u + 6)^{-1}du\]

Well, okay... yes, I did learn in calculus II about the root technique thingy

then it'll just be integration of 1/u

\[ \int {1 \over u}du\]\[\ln|u|\]\[\ln|x^2 +6|\]?

Oh well, nope.

I turned \(dx\) to \(du\).

oops

\[u = x^2 + 6\]\[du = 2xdx\]

nope. that was the ryt answer. that + C.

Protect me from Calculus, O Heaven!

your answer was correct ln |x^2+6| + c

\[\int {1 \over u}du\]\[\ln|x^2 + 6| + C\]I always slip the + c off.

I need more practice. Ugh!

parth u learning limits also ??

Already learnt, yes.

ok, i was gonna give a tutorial on that....and definite integration...which u want first ?

I'd like a tutorial on the applications of definite integrals!

*noted

....wow

... wow * 2

That's @TuringTest's intuition too!

but remember @Kainui i did not put in chain rule in what i wrote

this is differential calculus....how the heck did integration by parts come in o.O