$$\begin{align*} \boxed{\int 2x \ln(x) \, dx = x^2 \ln(x) – \frac{x^2}{2} + C} \end{align*}$$

where $C$ is the constant of integration.

Introduction

In calculus, integrating functions that combine both polynomial and logarithmic terms, like $2x \ln(x)$, requires the specific technique of integration by parts. This method is particularly useful for integrating products of different types of functions, such as polynomials and logarithms.

Step-by-Step Solution

To integrate $2x \ln(x)$, we will use the integration by parts formula:

$$\begin{align*} \int u \, dv = u v – \int v \, du \end{align*}$$

For our integral $\displaystyle \int 2x \ln(x) \, dx$:
– Let $u = \ln(x)$, so $du = \dfrac{1}{x} \, dx$
– Let $dv = 2x \, dx$, so $v = x^2$

This setup allows us to apply the integration by parts formula effectively. Its important to know which to set as $u$ and which as $dv$. In general, if it is easy to take the derivative, you should set that as $u$, while if it is easy to integrate, you should set that as $dv$.

Apply the Integration by Parts Formula

Now substitute into the formula $\displaystyle \int u \, dv = u v – \int v \, du$:

$$\begin{align*} \int 2x \ln(x) \, dx = \ln(x) \cdot x^2 – \int x^2 \cdot \frac{1}{x} \, dx \end{align*}$$

Simplify inside the integral:

$$\begin{align*} = x^2 \ln(x) – \int x \, dx \end{align*}$$

Step 3: Integrate $x$ with Respect to $x$

Now we integrate $x$:

$$\begin{align*} \int x \, dx = \frac{x^2}{2} \end{align*}$$

So our expression becomes:

$$\begin{align*} x^2 \ln(x) – \frac{x^2}{2} + C \end{align*}$$

where $C$ is the constant of integration.

Conclusion

To summarize, the integral of $2x \ln(x)$ is $x^2 \ln(x) – \dfrac{x^2}{2} + C$. By using integration by parts, we were able to handle the product of the polynomial and logarithmic terms effectively as long as we correctly identified which should be integrated, and which would be easier differentiated.

$$\begin{align*} \int 2x \ln(x) \, dx = x^2 \ln(x) – \frac{x^2}{2} + C \end{align*}$$