- #1
azatkgz
- 186
- 0
I'm not sure about answer.It looks very strange.
[tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}[/tex]
for u=lnx-->u'=1/x
[tex]\int \frac{du}{\sqrt{1+u^2}}[/tex]
substituting [tex]u=tan\theta[/tex]
[tex]=\int \frac{d\theta}{cos\theta}=ln|sec\theta+tan\theta|[/tex]
[tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}=ln|\sqrt{-1}|[/tex]
Homework Statement
[tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}[/tex]
The Attempt at a Solution
for u=lnx-->u'=1/x
[tex]\int \frac{du}{\sqrt{1+u^2}}[/tex]
substituting [tex]u=tan\theta[/tex]
[tex]=\int \frac{d\theta}{cos\theta}=ln|sec\theta+tan\theta|[/tex]
[tex]\int_{1}^{e}\frac{dx}{x\sqrt{1+ln^2x}}=ln|\sqrt{-1}|[/tex]
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