$$ \int ( \ln x ) ^ { r } d x = $$

Question

UpStudy AI · Accepted Answer

To solve the integral $$ \int (\ln x)^r \, dx $$, we can use integration by parts. Let's set:

- $$ u = (\ln x)^r $$ 
- $$ dv = dx $$

Then, we differentiate and integrate:

- $$ du = r (\ln x)^{r-1} \cdot \frac{1}{x} \, dx $$
- $$ v = x $$

Now, applying the integration by parts formula $$ \int u \, dv = uv - \int v \, du $$:

$$
\int (\ln x)^r \, dx = x (\ln x)^r - \int x \cdot r (\ln x)^{r-1} \cdot \frac{1}{x} \, dx
$$

This simplifies to:

$$
\int (\ln x)^r \, dx = x (\ln x)^r - r \int (\ln x)^{r-1} \, dx
$$

Thus, we have:

$$
\int (\ln x)^r \, dx = x (\ln x)^r - r \int (\ln x)^{r-1} \, dx
$$

This result can be used recursively to express the integral in terms of lower powers of $$ \ln x $$.

For example, if you want to compute $$ \int (\ln x)^2 \, dx $$, you would use:

$$
\int (\ln x)^2 \, dx = x (\ln x)^2 - 2 \int \ln x \, dx
$$

And you would need to compute $$ \int \ln x \, dx $$ separately, which is:

$$
\int \ln x \, dx = x \ln x - x + C
$$

In summary, the integral $$ \int (\ln x)^r \, dx $$ can be expressed as:

$$
\int (\ln x)^r \, dx = x (\ln x)^r - r \int (\ln x)^{r-1} \, dx
$$

This formula allows you to compute the integral for any non-negative integer $$ r $$ recursively.
The integral $$ \int (\ln x)^r \, dx $$ can be expressed as $$ x (\ln x)^r - r \int (\ln x)^{r-1} \, dx $$.

\( \int ( \ln x ) ^ { r } d x = \)

Real Tutor Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Calculus Questions