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

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

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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 | 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