1. El_Arrow

2. El_Arrow

i understand the completing the square part i dont understand where the tangent came from

3. El_Arrow

@freckles @jim_thompson5910 @Zarkon

4. El_Arrow

hello?

5. freckles

$\int\limits_{}^{}\frac{1}{u^2+1} du =\arctan(u)+C \\ \text{ and you have } \frac{1}{4} \int\limits \frac{dx}{\frac{1}{4}(x+3)^2+1} =\frac{1}{4} \int\limits \frac{dx}{(\frac{1}{2}(x+3))^2+1} \\ \text{ Let } u=\frac{1}{2}(x+3) \\ du=\frac{1}{2} dx$

6. 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

7. yolomcswagginsggg

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8. El_Arrow

@freckles how did you get the 1/2 out front of (x+3)^2?

9. freckles

$a+b=b(\frac{a}{b}+1) \\ \text{ so like here we had } \\ (x+3)^2+4 =4[\frac{1}{4}(x+3)^2+1]$

10. freckles

I just factored a 4 out on the bottom

11. El_Arrow

so (x+3)^2 is a and the 4 is b correct?

12. freckles

$(x+3)^2+4 \\ = \\ \frac{4}{4}(x+3)^2+4 \\ =4 \cdot \frac{1}{4}(x+3)^2+4 \cdot 1 \\ = \\ =4[\frac{1}{4}(x+3)^2+1]$

13. freckles

yeah in that example yes you can say that which is perfectly fine

14. freckles

I put a few more lines in between there maybe it to make it more clear hopefully

15. El_Arrow

okay thanks @freckles

16. freckles

np