Question

Evaluate the integral from 1sqrt(3) to sqrt(3) of 8/(1+x^2) dx

Answer 1

indefinite integral is 8 inverse tan of x

Answer 2

@313swagg As your question is currently written, the answer is \(0\) since the bounds appear to be the exact same place.

Answer 3

yeah I know that much but how do you do the limit from 1sqrt(3) to sqrt(3)

Answer 4

answer in back of the book is 4pi/3

Answer 5

\(1\sqrt3=\sqrt3\). What do you mean by "1sqrt3".

Answer 6

\[\int\limits_{1\sqrt{3}}^{\sqrt{3}} 8/(1+x^2) dx\]

Answer 7

\[\int_\frac{1}{\sqrt{3}}^{\sqrt{3}}\frac{8}{1+x^2}dx\]

Answer 8

just \[1\sqrt{3}\] not1 over sqrt3

Answer 9

The question, as formulated literally by @313swagg , has an answer of \(0\) because \(\int_a^af(x)\,dx=0\).

Answer 10

i am sure no one write
\[1\sqrt{3}\] it may be bad typesetting

Answer 11

okay thats probably a mistake in the book then but still how do i evaluate this if it is 1/sqrt3

Answer 12

\[8\left(\tan^{-1}(\sqrt{3})-\tan^{-1}(\frac{1}{\sqrt{3}})\right)\] is probably what you need

Answer 13

okay yeah that makes sense so did you have to do the chain rule for this and substitution?

Answer 14

\(\arctan(\frac{1}{\sqrt{3}})=\frac{\pi}{6}\)

Answer 15

neither. you need to recall that \(\frac{d}{dx}\tan^{-1}(x)=\frac{1}{x^2+1}\)

Answer 16

ok that makes sense thank you!