How do you evaluate the definite integral #int (dx/(xsqrt(lnx)))# from #[1, e^6]#?

Question

Emilia Aguilar · Accepted Answer

#2sqrt6#

Begin by using the substitution #u = ln(x)#.

From this we know that #du = 1/xdx#

Also the limits of our integral will change to: #1: ln(1) = 0# and #e^6 : lne^6 = 6#

Thus we can substitute this into the integral as follows:

#int_1^(e^6)1/(xsqrt(lnx))dx=int_0^6 1/sqrt(u)du#

Now evaluate the integral:

#=int_0^6u^(-1/2)du=[2u^(1/2)]_0^6=2sqrt(6)#

Elijah Abram · Answer

undefined To evaluate the definite integral $\int_{1}^{e^6} \frac{dx}{x \sqrt{\ln x}}$, we can use the substitution method. Let $u = \ln x$, then $du = \frac{1}{x} dx$. This transforms the integral into:

$$
\int \frac{dx}{x \sqrt{\ln x}} = \int \frac{du}{\sqrt{u}}
$$

Now, integrating $\frac{du}{\sqrt{u}}$ with respect to $u$ gives $2\sqrt{u} + C$, where $C$ is the constant of integration. Substituting back $u = \ln x$, we get $2\sqrt{\ln x} + C$.

Now, evaluating the definite integral from $1$ to $e^6$ gives:

$$
\left[2\sqrt{\ln x}\right]_{1}^{e^6} = 2\sqrt{\ln(e^6)} - 2\sqrt{\ln(1)}
$$

Since $\ln(e^6) = 6$ and $\ln(1) = 0$, we have:

$$
2\sqrt{6} - 0 = 2\sqrt{6}
$$

So, the value of the definite integral $\int_{1}^{e^6} \frac{dx}{x \sqrt{\ln x}}$ is $2\sqrt{6}$.