Answer to Integral Calculus example. Here is a photo

Ryan Saladores

Answered question

2022-07-23

karton

Expert2023-06-02Added 613 answers

To solve the integral

\int \frac{4^{\ln \frac{1}{x}}}{x} d x

, we can simplify the expression using logarithmic properties.
Let's start by rewriting

4^{\ln \frac{1}{x}}

4^{\ln \frac{1}{x}} = e^{\ln 4^{\ln \frac{1}{x}}} = e^{\ln \frac{1}{x} \cdot \ln 4}

Next, we can rewrite the integral as:

\int \frac{4^{\ln \frac{1}{x}}}{x} d x = \int \frac{e^{\ln \frac{1}{x} \cdot \ln 4}}{x} d x

Now, we can simplify further by bringing the exponent inside the integral:

\int \frac{e^{\ln \frac{1}{x} \cdot \ln 4}}{x} d x = \int \frac{e^{\ln 4 \cdot \ln \frac{1}{x}}}{x} d x

Using the property

e^{\ln a} = a

, we have:

\int \frac{e^{\ln 4 \cdot \ln \frac{1}{x}}}{x} d x = \int \frac{4 \cdot \ln \frac{1}{x}}{x} d x

We can now separate the terms and rewrite the integral as:

\int \frac{4 \cdot \ln \frac{1}{x}}{x} d x = 4 \int \frac{\ln \frac{1}{x}}{x} d x

To proceed with the integration, we can use the substitution method. Let's substitute

u = \ln \frac{1}{x}

\frac{d u}{d x} = - \frac{1}{x} ⟹ d x = - \frac{d u}{u}

Substituting into the integral:

4 \int \frac{\ln \frac{1}{x}}{x} d x = 4 \int \frac{u}{- u} d u = - 4 \int d u = - 4 u + C

Finally, substituting back

u = \ln \frac{1}{x}

- 4 u + C = - 4 \ln \frac{1}{x} + C

Therefore, the solution to the integral

\int \frac{4^{\ln \frac{1}{x}}}{x} d x

- 4 \ln \frac{1}{x} + C

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?