Integration By Parts(stuck on one part)

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Homework Statement



Evaluate the Definite Integral
Integral[0,1] Ln(x^2+1) dx

Homework Equations





The Attempt at a Solution


So far I have done the first part,
U= Ln (x^2+1) Du= 2x/(x^2+1)dx
Dv= 1 dx V=x

x* Ln(x^2+1)- Integral [0,1] (2x^2/(x^2+1) dx

But I'm stuck on the second integral, I have tried punching into a TI-89, but It comes out to a weird answer. Can someone help?
 
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Start by dividing the polynomial x^2+1 into x^2 to express it as quotient and remainder.
 
What do I do after that? How does that help?
 
x^2/(1+x^2)=1-1/(1+x^2). You can integrate 1 and you should be able to integrate 1/(1+x^2). It's an inverse trig function.
 
Oh wow, thanks!
:smile:
 
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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