How to Integrate 6t(4+t^2)^.5 dt using u-substitution?

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Homework Statement



6t(4+t^2)^.5 dt




The Attempt at a Solution



I know the answer...what are the steps of how to get there
2(4+t^2)^3/2
 
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rjs123 said:

Homework Statement



6t(4+t^2)^.5 dt

The Attempt at a Solution



I know the answer...what are the steps of how to get there
2(4+t^2)^3/2
Do you know how integration by substitution is done? If not, look it up in your book.
Try a u-substitution. That is let u=4+t^2. Where do you go from here?
 
sutupidmath said:
Do you know how integration by substitution is done? If not, look it up in your book.
Try a u-substitution. That is let u=4+t^2. Where do you go from here?

du = 2t dt

dt = du/2t

3[u^.5 * du]

3(u^1.5)/1.5

2 (4 + t^2)^1.5

is this how it should be done?
 
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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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