x.lnx2 integral | What is integral of x.lnx^2? | ∫x.lnx2.dx=?

 x.lnx2 integral | What is integral of x.lnx^2? | ∫x.lnx2.dx=?

Integral of x * ln(x²) | What is the Integral of x * ln(x²)?

In calculus, integrals play a significant role in finding areas under curves, among other applications. In this article, we will explore the process of solving the integral of the function ( x \ln(x^2) ).

Understanding the Function

The function we want to integrate is ( x \ln(x^2) ). First, let's simplify it using properties of logarithms:

$\ln(x^2) = 2 \ln(x)$

This transforms our integral to:

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

Now we focus on finding the integral of ( x \ln(x) ).

Integration by Parts

To calculate ( \int x \ln(x) , dx ), we will use the method of integration by parts. The formula for integration by parts is:

$\int u , dv = uv - \int v , du$

Let’s set:

( u = \ln(x) ) --> which implies ( du = \frac{1}{x} , dx )

( dv = x , dx ) --> which gives us ( v = \frac{x^2}{2} )

Now we can substitute into the integration by parts formula:

$\int x \ln(x) , dx = \ln(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} , dx$

This simplifies to:

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

Next, we compute ( \int \frac{x}{2} , dx ):

$\int \frac{x}{2} , dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}$

Compiling the Results

Putting all of the pieces together, we have:

$\int x \ln(x) , dx = \frac{x^2}{2} \ln(x) - \frac{x^2}{4} + C$

where ( C ) is the constant of integration.

Final Result

Now, we can return to our original integral:

$\int x \ln(x^2) , dx = 2 \int x \ln(x) , dx = 2 \left( \frac{x^2}{2} \ln(x) - \frac{x^2}{4} \right) + C$

This simplifies to:

$\int x \ln(x^2) , dx = x^2 \ln(x) - \frac{x^2}{2} + C$

Conclusion

Thus, the integral of the function ( x \ln(x^2) ) is given by:

$\int x \ln(x^2) , dx = x^2 \ln(x) - \frac{x^2}{2} + C$

This result showcases the beauty of calculus and the power of integration by parts, allowing us to tackle more complex functions systematically. Understanding these techniques extends our ability to solve a vast array of calculus problems efficiently.