Question

solve dy/dx = x/(x^2 +1)

Answer 1

Hi
\[dy=x/x^2+1dx \rightarrow y=\ln(x^2+1)/2+c\]

Answer 2

quotient rule @suediff
It might help if you think of it like this (learned this when I took this course):

low D high - high D low all over (low low).

What that means IS: you take the lower function (numerator) and multiply it by the high (numerator), then subtract the higher function (numerator) and multiply it by the derivative of the denominator.
Then, divide everything by the denominator squared, hence why I have low low :)

hope this helps.

Answer 3

\[f=\frac{u}{v}\\f'=\frac{u'v-v'u}{u^2}\]

Answer 4

dy=intxdx/x^2+1
dy=1/2 int( 2x dx)/(x^2+1)
y =1/2 log(x^2+1) +c