Stuck on the integral: arctan (4t) dt

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Homework Statement



The Integral is arctan (4t) dt


Homework Equations



I know how to do integration by parts, but I guess I have forgotten some of the integration rules.

The Attempt at a Solution



I set ∫arctan(4t)=u, and dt=dv

I know that the derivative of arctan(x) is 1/(1+x^2), But when I differentiate arctan(4t), it comes out as 4t/(1+16t^2). Why is this? To me it seems like it should be 1/(1+4t^2). I know how to do the rest, I have the answer, I'm just not sure how they got there. Thanks for any help.
 
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dlikes said:
But when I differentiate arctan(4t), it comes out as 4t/(1+16t^2). Why is this?

Because you've calculated the derivative wrong. To tell you exactly where your error is, you'll need to post your steps.
 
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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