Having trouble with the antiderivative of this function

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SUMMARY

The discussion focuses on evaluating the indefinite integral of the function \(\int \frac{7x + 1}{x^2 + 1} dx\). The initial approach involved substitution with \(u = x^2 + 1\), leading to a transformation of the integral. The solution was clarified by splitting the integral into two parts: \(\int \frac{7x}{x^2 + 1} dx\) and \(\int \frac{1}{x^2 + 1} dx\). The final answer is confirmed as \(\frac{7}{2} \ln(x^2 + 1) + \arctan(x)\).

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Homework Statement


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

Homework Equations


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.
 

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