Question

find integrate (1/(xsqrt(1+(lnx)^2)),1,e,x)

Answer 1

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

Answer 2

\[\int\limits_{1}^{e}\frac{dx}{x \sqrt{1+(\ln(x))^2}}\]
\[\text{ Let } a=\ln(x) => da=\frac{dx}{x}\]
if x=e then a=ln(e)=1
if x=1 then a=ln(1)=0
So we have
\[\int\limits_{0}^{1}\frac{da}{\sqrt{1+a^2}} \]

Answer 3

ln x= t .. then this becomes dt/(sqrt(1+t^2)) .. now put t = tany and you get secydy = ln(secy + tany)

Answer 4

\[\text{ Let } a=\tan(\theta) => da=\sec^2(\theta) d \theta\]
if a=1 then theta=arctan(1)=pi/4
if a=0 then theta=arctan(0)=0
\[\int\limits_{0}^{\frac{\pi}{4}}\frac{\sec^2(\theta) d \theta}{\sqrt{1+\tan^2(\theta)}}\]

Answer 5

yea thats as far as i got and lost

Answer 6

\[\int\limits_{0}^{\frac{\pi}{4}}\sec(\theta) d \theta \text{ (note :} 1+\tan^2(\theta) =\sec^2(\theta) \text{ )}\]

Answer 7

\[=\ln|\sec(\theta)+\tan(\theta)||_0^\frac{\pi}{4}=\ln|\sec(\frac{\pi}{4})+\tan(\frac{\pi}{4})|-\ln|\sec(0)+\tan(0)|\]

Answer 8

\[=\ln(\frac{1}{\frac{\sqrt{2}}{2}}+\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}})-\ln(\frac{1}{1}+\frac{0}{1})\]

Answer 9

\[=\ln(\frac{2}{\sqrt{2}}+1)-\ln(1)=\ln(\frac{2 \sqrt{2}}{2}+1)-0=\ln(\sqrt{2}+1)\]

Answer 10

for the absolute values, how do you get rid of it?

Answer 11

thanks for your help! very clear instruction. i always liked your help =)

Answer 12

the number inside was positive you don't need them anymore

Answer 13

\[\ln|1|=\ln(1)\]
You don't need the absolute value since |1|=1