Integral of x(ln x)^4: Steps & Solution

Quadruple Bypass
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Does the integral of x(ln x)^4 =
x^2/x(ln x)^4 - x^2(ln x)^3 + 3/2 x^2 (ln x)^2 - 3/2 x^2(ln x) + 3/2 x +C ?

Or did I do something completely wrong?
Sorry I didn't show my work, it would probably take me 30 mins to type it up here.
 
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Why don't you try differentiating it and see? That's all we could do anyway, and we won't do your tedious work for you.
 
Maple gives

1/2*x^2*ln(x)^4-x^2*ln(x)^3+3/2*x^2*ln(x)^2-3/2*x^2*ln(x)+3/4*x^2+C

and, in the future, try www.integrals.com (which in this case does not give the form of the answer your looking for.)
 
LOL, i forgot about differentiating that. Damn, i was so tired after solving that my mind shut down. LOL THanks.

Thanks benorin too, my professor was telling us about some website that does that, and i forgot to write it down.
 
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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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