Question

If
f(x) = int_{2}^{x} 1/{1+t^5} dt

then f'(x) =

Answer 1

recall that the integral is also called an antiderivative

Answer 2

\[\frac{ d }{ dx }\int\limits x^2 dx = x^2 \]

Answer 3

oh is that an integral with bounds of 2 to x of 1/(1+t^5)dt

Answer 4

yeah thats it

Answer 5

\[\int_2^x\frac{1}{1+t^5}dt\]

Answer 6

d/dx integral {a,x} f(t) = f(x)

Answer 7

So the opposite of this is also true. \[d(\tan^{-1}(x)) = \frac{1}{1+x^2} \implies \int\frac{1}{1+x^2}dx = \tan^{-1}(x)\]

Answer 8

Sorry, I am missing something. \[\int\frac{du}{a^2+u^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C\]

Answer 9

you want to use the fundamental theorem of calculus here