i completed the square but i didnt get the same form as you.

i got.. (x + 1/2)^2 + 11/4



im worried, if these type of questions will be on the exam where u have to manipulate the form into something more recongizeable

This post was edited by FamilyGuyViewer on Sep 4 2016 04:33pm

Yes, (x+1/2)² + 11/4

But the constant 11/4 is not so welcomed, since you want to use ∫ dt / (t²+1)

Just factorize by 11/4 :

(x+1/2)² + 11/4 = 11/4 [ (x+1/2)² * 4/11 + 1 ]

= 11/4 * [ ( (x+1/2)(2/√11) )² + 1 ]

= 11/4 * [ ( (2x+1)/√11 )² + 1 ]

not sure how u got this step.. i know u multiplied everyhting by 4/11 but then wat

also, the 2nd line.. not sure how u got the 2/sqrt 11

This post was edited by FamilyGuyViewer on Sep 4 2016 06:33pm

Your work seems fine.

(4/11) ∫ √11 du / (u²+1)

is correct.

Just finish like this :
√11 / 11 = 1 / √11 of course, then :

(4/√11) ∫ du / (u²+1) = (4/√11) . Arctan (u)

and since u = (2x+1)/√11 :

(4/√11) . Arctan ( (2x+1)/√11 )

thanks man! u da best! ur a fckin genius

http://www.wolframalpha.com/input/?i=integrate+%5B(1+-+2x)+%2F+(x%5E2+%2B+x+%2B3)%5D+dx

how come the result from wolframalpha has a log term in the result??

Same with ∫ [(1 - x) / (x^2 + x +3)] dx :

(1 - x) / (x² + x + 3) = - (1/2) * ( 2x + 1) / (x² + x + 3) + (3/2) / (x² + x + 3)

The idea is the same, you want the derivative of (x² + x + 3) as a numerator, so you just write it... and adjust to keep it right.



The log comes from the integral of the first term : ∫ u' / u = log |u| + constant.