Evaluate the definite integral:
int_{1}^{e^9} \frac{dx}{x \sqrt{\ln x}} =
i gt my answer as 1^(e^9) but it's saying it's wrong
To evaluate the definite integral, you need to compute the antiderivative of the function and apply the Fundamental Theorem of Calculus to find the definite integral.
Let's break down the steps:
Step 1: Find the antiderivative of the function
The given function is \(\frac{1}{x \sqrt{\ln x}}\).
To find the antiderivative, we can set \(u = \ln x\) and find the antiderivative of \(\frac{1}{\sqrt{u}}\) with respect to \(u\).
Using the substitution method, let \(u = \ln x\), then \(du = \frac{1}{x} dx\).
Hence, the integral can be rewritten as:
\(\int \frac{dx}{x \sqrt{\ln x}} = \int \frac{du}{\sqrt{u}}\).
Integrating \(\frac{1}{\sqrt{u}}\) gives us \(2\sqrt{u} + C\), where \(C\) is the constant of integration.
Step 2: Evaluate the definite integral using the Fundamental Theorem of Calculus
Applying the Fundamental Theorem of Calculus, the definite integral is given by:
\(\int_{1}^{e^9} \frac{dx}{x \sqrt{\ln x}} = [2\sqrt{\ln x}]_{1}^{e^9}\).
Evaluate the expression by plugging in the limits of integration:
\([2\sqrt{\ln (e^9)}] - [2\sqrt{\ln(1)}]\).
Simplifying, we have:
\([2\sqrt{9\ln e}] - [2\sqrt{0}]\).
Since \(\ln e = 1\) and \(\sqrt{0} = 0\), the expression becomes:
\[2\sqrt{9} - 0 = 2 \cdot 3 = 6\].
Therefore, the correct answer is 6, not 1 raised to the power of \(e^9\).