Find the value of Int[ (z+1)/ ( 3z^3+6z +5)^(1/3) dz]

2 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

int [ (z+1)/(3z^2 + 6z + 5)^1/3  dz

Let t =( 3z^2 + 6z + 5)^1/3

==> dt = (1/3)*(6z+6) (3z^2 + 6z + 5)^-2/3  dz

             = (1/3)*6(z+1)/ (3x^2 + 6x + 5)^2/3

              = 2(z+1)/ t^2  dz

==> dz = dt/2(z+1) t^2

==: int [(z+1)/(3z^2+6z + 5)^1/3 dz

   = intg[ (z+1)/ t  dt/2(z+1)t^2

   = int [1/2t^3  dt]

   = int (1/2)t^-3 dt

     = (1/2)t^-2/-2  + C

    = -(1/4)t^-2  + C

     =( -1/4t^2 + C

 = -1/4(3z^2 + 6z + 5)^2/3  + C

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We need to find the value of Int [(z+1) / (3z^2 + 6z +5) ^ (1/3) dz]

First we substitute 3z^2 + 6z +5 = u

=> du/dz = 6z +6 = 6*(z+1)

=> du = 6*(z+1) dz

Therefore we can substitute [(z+1) / (3z^2 + 6z +5) ^ (1/3) dz] with (1/6)/ u^ (1/3) du

=> Int [(z+1) / (3z^2 + 6z +5) ^ (1/3) dz]

=> Int [(1/6) / u^ (1/3) du]

=> (1/6) Int [u^ (-1/3) du]

=> (1/6) [u^ (-1/3 +1) / (-1/3 +1)] +C

=> (1/6) *u^ (2/3)/ (2/3) +C

=> (1/6)*(3/2) u^ (2/3) +C

=> (1/4) u^ (2/3) +C

=> (1/4) (3z^2 + 6z +5) ^ (2/3) +C

Therefore the required result is (1/4) (3z^2 + 6z +5) ^ (2/3) +C

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question