Answer:
[tex]ln(2)x+\frac{1}{2} x^2ln(x)-\frac{1}{4} x^2+C[/tex]
Step-by-step explanation:
We begin with the integral: [tex]\int {(ln(2x^x)} \, dx[/tex]
To integrate this, we need to apply a few log properties to separate this, First, we can use the product rule to separate these two terms. This give us
[tex]\int ln(2)+ln(x^x)dx[/tex]
We can then apply the power rule to the second log to get
[tex]\int ln(2)+xln(x)dx[/tex]
Now, we can separate these two integrals
[tex]\int ln(2)dx+\int xln(x)dx[/tex]
Let us solve the first integral. [tex]\int ln(2)dx =ln(2)x+C[/tex]
Now we can move on to the second integral: [tex]\int xln(x)dx[/tex]
To integrate this, we can use integration by parts.
[tex]u=ln(x)\\du=\frac{1}{x}dx \\\\dv=xdx\\v=\frac{1}{2} x^2[/tex]
Using the formula for integration by parts, this gives us:
[tex]\frac{1}{2} x^2ln(x)-\int{\frac{1}{x}*\frac{1}{2}x^2 } \, dx \\\\\frac{1}{2} x^2ln(x)-\int\frac{1}{2}x } \, dx\\\\\frac{1}{2} x^2ln(x)-\frac{1}{4} x^2+C[/tex]
Now that we have integrated both part, we can add them together and combine their C values to get our answer:
[tex]ln(2)x+\frac{1}{2} x^2ln(x)-\frac{1}{4} x^2+C[/tex]
