# Integral of 1/1+25x^2

## Homework Statement

integral of 1/1+25x^2 evaluated at sqrt.3/5 and 0/

## Homework Equations

arctan=1/1+x^2
arcsing=1/(sqrt.(1-x^2))
lnx=1/x

## The Attempt at a Solution

not sure if i have to use substitution or use lnx=1/x

this is what i tried, probably not right, integral 1/1+25x^2 = ln(1+25x^2)

## Answers and Replies

You were on the right track with that first relevant equation, arctan=1/1+x^2. I'd go with that.

Your answer of ln(1+25x^2) does not work, since taking the derivative will end up giving you a 50x on top, which doesn't match what you integrated.

thanks.

so: ddx arctan= 1/1+x^2

so i think i have to use substitution to get the answer.

so i can rewrite the equation as 1/1+(5x)^2

u= 5x
du= 5 dx
dx = du/5

then i would get 1/2 du / 1+u^2 which i could then make into 1/5arctan(u)

then the integrals would have to be changed x=0, u=0, x=(sqrt. 3) / 5, u= sqrt. 3

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