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.

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