- #1

- 190

- 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]

Last edited: