find the integral S 9z square root 3z^2-7 dz

Question

find the integral S  9z square root  3z^2-7 dz

anonymous · Answer

First guess, your u is under the radical. Good indication your guess is right, du is outside radical.

anonymous · Answer

do i divide 9 by du?

anonymous · Answer

Divide?! Do not pass go. Go back to square one. Are you familiar with integrating using u substitution method.

anonymous · Answer

this is what i have$$u: 3z ^{2}-7  du: 1/3  $$

anonymous · Answer

$$^9\int\limits_{}^{}\sqrt{u} = 1/3du = 9^{1/2}= 9u ^{3/2}/3/2 \times  1/3du  =6u3/2 \times 1/3 =2(3z ^{2}-7)^3/2 $$

anonymous · Answer

du=6z dz
du/6=z
Don't concern yourself with trying to cancel the 9. It is just a constant and you can bring it front of the integral sign. It seems like you have been distracted by the 9 and did not get the correct derivative of u.

anonymous · Answer

oooh i see  lol