Integration by Parts: Solve y(1+y^2)^1/2 dy

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integral y(1+y^2)^1/2 dy

can someone help me with this?

Thanks.
 
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Wouldn't this solve easier with u substituion?
u = 1+y^2
du = 2y dy

Solution is 1/3(1+y^2)^3/2

If you want integration by parts, multiply it out and make u to be y so that du is 1.

By trig you'll need to substitute tan because the problem is + Y^2.

Bernie
 
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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