Question

How can I finish this integral?

My Result of integral:
x(ln^2x+4lnx+1) is
(3x^2lnx)+(x^2)/(2)-(x^3lnx)/(3)-(5x^3)/(9)
But when I do the limit (from e^-2-sqrt3 to -2+sqrt3) I can't get result 123sqrt3 .
What I am doing wrong?😕

Answer 1

Can someone help?

Answer 2

You could group your integral this way
\[ \int (x \ln^2 x + x \ln x)+(2 x \ln x + x) + x \ln x \ dx\]

the terms in parens are "exact" integrals (up to a factor):
\[ \int \frac{1}{2}\left(2x \ln^2 x + 2x \ln x\right)+(2 x \ln x + x) + x \ln x \ dx\]
i.e
\[ \int \frac{1}{2} d(x^2 \ln^2 x) + d(x^2 \ln x) + x \ln x \ dx\]
thus we get
\[ \frac{1}{2} x^2 \ln^2 x + x^2 \ln x + \int x \ln x \ dx\]
we can use integration by parts to evaluate the last integral
(let u = ln x and dv = x dx)
the last integral gives us \[ \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 \]
combining like terms
\[ \frac{1}{2} x^2 \ln^2 x + \frac{3}{2} x^2 \ln x- \frac{1}{4} x^2 \]

However, I can't make out what the limits are

Answer 3

Limits are from e^(-2-sqrt3) to (-2+sqrt3).

Answer 4

Not that this makes things much easier, but you could alternatively substitute \(\ln x=u\) so that \(e^u=x\) and \(e^u\,\mathrm{d}u=\mathrm{d}x\). Then
\[\int_{e^{-2-\sqrt3}}^{e^{-2+\sqrt3}}x\left(\ln^2x+4\ln x+1\right)\,\mathrm{d}x=\int_{-2-\sqrt3}^{-2+\sqrt3} e^{2u}\left(u^2+4u+1\right)\,\mathrm{d}u\](... if only to clean up those limits. I'm also assuming you meant the upper limit to be the argument to \(\exp\)?) The resulting integral is, like its previous form, a perfect candidate for integration by parts.