Having trouble with the antiderivative of this function

jlt90
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Homework Statement


Evaluate the indefinite integral:
\int7x+1/x^2+1 dx

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The Attempt at a Solution


My first attempt at the solution was to try using substitution. I set u=x^2+1. so du=2x dx and x=sqrt(u-1). Then I rewrote the integral so it is \int7du/4usqrt(u-1). This is where I don't know where to go with this attempt.

I'm pretty sure I'm going about this problem all wrong. If I could just get a push in the right direction I'd really appreciate it :)
 
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Split the integral into 7x/(x^2+1) and 1/(x^2+1). The first one is the u-substitution you spoke of. The second is a trig substitution.
 
Thank you very much. So the first one becomes \int7/2u du then 7/2ln(x^2+1) after taking the anti derivative and substituting x^2+1 for u. Then the second part becomes arctan x. So would the final answer be (7/2)ln(x^2+1)+arctan(x)?
 
That looks fine to me.
 
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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