# How do you integrate 1/(xlnx)dx?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

21
Feb 15, 2015

Answer : $\ln \ln x + c$

Details :

Basic Fact used:
This is an application of the fact that $\int \frac{\mathrm{dx}}{x} = \ln x + c$

Substitution :
Look at the form of the integral: $\frac{1}{\ln} x \frac{\mathrm{dx}}{x}$. Remembering the fact above, one can substitute $t = \ln x$ so that $\mathrm{dt} = \frac{\mathrm{dx}}{x}$.

As a result the integral simplifies to $\int \frac{1}{\ln} x \frac{\mathrm{dx}}{x} = \int \frac{\mathrm{dt}}{t} = \ln t + c$

Back Substitute $t = \ln x$ to get the answer.

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

16
vince Share
Feb 15, 2015

Hello !

I propose another solution.

Remember that $\left(\setminus \ln \left(u\right)\right) ' = \setminus \frac{u '}{u}$ if $u$ is a positive differentiable function.

Take $u \left(x\right) = \setminus \ln \left(x\right)$ for $x > 1$ : it's a positive differentiable function.

Remark that $\setminus \frac{u ' \left(x\right)}{u \left(x\right)} = \setminus \frac{\setminus \frac{1}{x}}{\setminus \ln \left(x\right)} = \setminus \frac{1}{x \setminus \ln \left(x\right)}$, then

$\setminus \int \setminus \frac{\setminus \textrm{d} x}{x \setminus \ln \left(x\right)} = \setminus \ln \left(u \left(x\right)\right) + c = \setminus \ln \left(\setminus \ln \left(x\right)\right) + c$,

where $c$ is a real constant.

• 12 minutes ago
• 13 minutes ago
• 14 minutes ago
• 17 minutes ago
• 38 seconds ago
• 3 minutes ago
• 9 minutes ago
• 9 minutes ago
• 11 minutes ago
• 11 minutes ago
• 12 minutes ago
• 13 minutes ago
• 14 minutes ago
• 17 minutes ago