An integration involving inverse trig.

Charismaztex
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Homework Statement



The question is integrate

\int_0^1 (2+\frac{1}{x+1}+\frac{2x+1}{x^2+4}) dx

Homework Equations



\int_0^1 (2+\frac{1}{x+1}+\frac{2x+1}{x^2+4}) dx =2 + ln(\frac{5}{2}) +\frac{1}{2}arctan(\frac{1}{2}) (yes, it's a prove question)

The Attempt at a Solution



The first two parts I have no problem but I am not quite sure how to integrate the

\frac{2x+1}{x^2+4} part. I know it has to do with arctan but I'm not quite sure about the 2x+1 part.

Thanks in advance,
Charismaztex
 
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Split up the last fraction so that you have \frac{2x}{x^{2}+4} + \frac{1}{x^{2}+4}. Use a simple substitution for the first fraction and then the second fraction will involve the inverse tangent. In general,

\int \frac{dx}{x^2+a^2} = \frac{1}{a} arctan(\frac{x}{a}) + C
 
ahh! I completely missed that step! Thanks for your help :)

Charismaztex
 
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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