# Are you calculus-literate?

I need help with the following integral:

7. (6t + 2) / (t^2 + 2t + 17)

I understand that one needs to complete-the-square with the denominator and then use a substitution. I am confused as to how to go about the latter part. Any help is much appreciated.

Relevance

int [ (6t + 2) / (t^2 + 2t + 17) dt ]

= int [ (6t + 2) / [ (t + 1)^2 + 16] dt ]

Let u = t + 1

du = dt

= int [ (6u - 4) / (u^2 + 16) du ]

= 6 * int [ u / (u^2 + 16) du ] - int [ 4 / (u^2 + 16) ]

for the first integral: v = u^2 + 16; (1 / 2)dv = u du

= 6 * (1 / 2) * int [ (1 / v) dv ] - 4* int [ 1 / (u^2 + 16) du ]

= 3 ln abs ( v ) - 4*(1 / 4)arctan (u / 4) + C

= 3 ln abs (u^2 + 16) - arctan (u / 4) + C

= 3 ln abs [ (t + 1)^2 + 16 ] - arctan [ (t + 1) / 4 ] + C

• Anonymous

The denominator is (t + 1)^2 + 16

(6t + 2) / (t^2 + 2t + 17) = ( 6t + 2 ) / ( (t + 1)^2 + 16 )

** t + 1 = 4u

( 6t + 2 ) / ( 16u^2 + 16 )

( 6t + 6 - 4 ) / ( 16u^2 + 16 )

( 6(t + 1) - 4 ) / ( 16u^2 + 16 )

( 6(4u) - 4 ) / ( 16u^2 + 16 )

( 24u - 4 ) / ( 16u^2 + 16 )

4*( 6u - 1 ) / 16*( u^2 + 1 )

(1/4) * ( 6u - 1 ) / ( u^2 + 1 )

(1/4) * [ 6u / ( u^2 + 1 ) - 1 / ( u^2 + 1 ) ]

Ok, we have ::

INTEGRAL (1/4) * [ 6u / ( u^2 + 1 ) - 1 / ( u^2 + 1 ) ] dt

** dt/du = 4

** dt = 4*du

INTEGRAL [ 6u / ( u^2 + 1 ) - 1 / ( u^2 + 1 ) ] du