Integration of (3x^2+1)^(3/2) - Step-by-Step Solution

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Homework Statement



\int x(3x^2+1)^\frac{3}{2}

Homework Equations



The Attempt at a Solution



u=3x^2+1

du=6x dx

\frac{dx}{6}=xdx

\int\frac{du}{6} u^\frac{3}{2}

\frac{1}{6}\int u^\frac{3}{2}du

\frac{1}{6} \frac{u^\frac{5}{2}}{\frac{5}{2}}

\frac{5}{12}u^\frac{5}{2}

\frac{5}{12}(3x^2+1)^\frac{5}{2}
 
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\frac{1}{6} \frac{u^\frac{5}{2}}{\frac{5}{2}} =/= \frac{5}{12}u^\frac{5}{2}
 
\frac{2}{30}u^\frac{5}{2} ?
 
Yeah looks right
 
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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