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

Thus we can substitute this into the integral as follows:

Now evaluate the integral:

By signing up, you agree to our Terms of Service and Privacy Policy

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}).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7