Sunday, October 2, 2011

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

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

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