Question

integration question

Answer 1

\[\int\limits_{3/-2}^{1/-2} dx/4x^{2}+12x+13\]

Answer 2

complete the square in the denominator and then trig substitute

Answer 3

i tried that but.. got stuck at the +13

Answer 4

got \[(x+2)^{2}+8\]

Answer 5

8 is not a square and if u factorise (x+2)^2 u dont get 4x^2....

Answer 6

factor that four out of the denom first\[\int_{-\frac32}^{-\frac12}\frac{dx}{4x^2+12x+13}=\frac14\int_{-\frac32}^{-\frac12}\frac{dx}{x^2+3x+\frac{13}4}\]\[=\frac14\int_{-\frac32}^{-\frac12}\frac{dx}{x^2+3x+\frac94+1}\]

Answer 7

\[=\frac14\int_{-\frac32}^{-\frac12}{dx\over(x+\frac32)^2+1}\]

Answer 8

how did u get 9/4+1? =/

Answer 9

13/4=(9/4)+1

completing the square on x^2+3x means adding a term of 9/4

you can save some trouble by noticing that 13/4=1+9/4 so we can make our constant term from the fraction we already have

Answer 10

can u continue plz

Answer 11

trig sub:\[x+\frac32=\tan\theta\implies dx=\sec^2\theta d\theta\]change bounds...

Answer 12

=/