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Integral with trig substitution. A bit stuck.... will post question.
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\[\int\limits_{}^{} (8x+6)/(x^2+1) dx\] I know that I need to to set u=arctan, but I'm stuck afterwards...
why not split the problem \[\int\limits \frac{8x}{x^2 +1} + \frac{6}{x^2 + 1} dx\]
then you have a log function + an arctan
so you are looking at \[ 4 \int\limits \frac{2x}{x^2 + 1} dx + 6 \int\limits \frac{1}{x^2 + 1} dx \]
split the problem as above and use the substitution rule of \[\int\limits \] linear function / quadratic
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ah ok that works :). Thank you both for the help... I just needed to split it up and it worked.
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