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integrate root (x^2 +9) + root(x^2-9) / root(x^4 - 81) dx \int \frac {\sqrt{x^2+9} + \sqrt{x^2-9}}\sqrt (x^4 - 81) how do you enter using latex?

hmm ill guess ill just do it like that then :o

integrate (x^2 + 4x +2) / x dx under completing square topic once again

lols just forgot the delta = 0 part xD ty

i wasnt looking for substitution method but thanks anyways :P

thanks i get it now :D answer my question in 3u section as well if you have time :spin:

du = 1 where x is needed on top so on top put u-4?????/just guessing this is under completing square topic though

integrate between 4 and 0 x / root(x+4) dx ty :D

equation: x^2 + 4cx/(1+c^2) +1 = 0 gradient of log(1+x^2) a x=c is y' = 2c / (1+c^2) Show that delta = -4 (1-c^2 / 1+c^2)^2 for this quadratic, and hence that the only tangents to y = log (1+x^2) which are mutually perpendicular are those at x= -1 and x= 1. skip the show part and help me...

pew

its cos i repped you. When you were at 0 (unknown quantity or something)

The equation of the tangent to y=xlogx at x=e is y = 2x-e. Find the distance from this tangent to the origin :)

does I(1/2sin2x)^2 = 1/8 I (1-cos4x) formula?? (haven't learnt this yet :haha:) i would try do Isinaxdx = -1/a cosax somehow just wna know if its a new formula :tongue:

lol thats not all xD check again

lol ty a few minor errors there but its ok :)

Use logs to help find the derivatives: y = [(x+1) root (x-1)] /(x+2) Take logs of both sides to diff y = x ^(logx) LaTeX PLEASE lol ty xD

as helpful as always :P

what happened there? =(log(x)-1) changed to 1/ log(x) -1 :confused:

Show that y = x/logx is a solution of the eqation dy/dx = (y/x) - (y/x)^2 Show that y = log(logx) is a solution of the equation x d^2y/dx^2 + x(dy/dx)^2 + dy/dx = 0 Thx :ninja: