El_Arrow
  • El_Arrow
don't understand how it got that answer please help
Mathematics
katieb
  • katieb
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El_Arrow
  • El_Arrow
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El_Arrow
  • El_Arrow
i understand the completing the square part i dont understand where the tangent came from
El_Arrow
  • El_Arrow

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El_Arrow
  • El_Arrow
hello?
freckles
  • freckles
\[\int\limits_{}^{}\frac{1}{u^2+1} du =\arctan(u)+C \\ \text{ and you have } \frac{1}{4} \int\limits \frac{dx}{\frac{1}{4}(x+3)^2+1} =\frac{1}{4} \int\limits \frac{dx}{(\frac{1}{2}(x+3))^2+1} \\ \text{ Let } u=\frac{1}{2}(x+3) \\ du=\frac{1}{2} dx\]
jim_thompson5910
  • jim_thompson5910
One way is to use a table of integrals (if you can recognize the form) http://integral-table.com/ look at equation (9) under the "Integrals of Rational Functions" section
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yolomcswagginsggg
  • yolomcswagginsggg
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El_Arrow
  • El_Arrow
@freckles how did you get the 1/2 out front of (x+3)^2?
freckles
  • freckles
\[a+b=b(\frac{a}{b}+1) \\ \text{ so like here we had } \\ (x+3)^2+4 =4[\frac{1}{4}(x+3)^2+1]\]
freckles
  • freckles
I just factored a 4 out on the bottom
El_Arrow
  • El_Arrow
so (x+3)^2 is a and the 4 is b correct?
freckles
  • freckles
\[(x+3)^2+4 \\ = \\ \frac{4}{4}(x+3)^2+4 \\ =4 \cdot \frac{1}{4}(x+3)^2+4 \cdot 1 \\ = \\ =4[\frac{1}{4}(x+3)^2+1]\]
freckles
  • freckles
yeah in that example yes you can say that which is perfectly fine
freckles
  • freckles
I put a few more lines in between there maybe it to make it more clear hopefully
El_Arrow
  • El_Arrow
okay thanks @freckles
freckles
  • freckles
np

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