X*ln(x) = 1/2x^2*ln(x) - 1/4*x^2 not sure how they got it

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Hello everyone!

I'm doing a line integral but I'm confused on the integration method used:

I got to integral 1 to 3 2*(x^4+xln(x) ); But they then have:
x^5/5 + 1/2x^2 * ln(x) - 1/4x^2 and I'm not sure how they got the last part, 1/2x^2 * ln(x) - 1/4x^2

Thanks!

u = x isn't what they did I don't think, and its been awhile so I'm rusty on integration XD
 
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The 'hard' part is the integral of x*ln(x). And they just integrated by parts.
 
ahh thank you
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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