Calculus u-substitution integration I have a mistake somewhere please help

[jolly sparrow]

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[frigid dust]

integral of ab is not integral of a times integral of b

[river elk]

then?

[frigid dust]

what usub can you do

[river elk]

9z

du = 9 dz

[frigid dust]

maybe (9z)^3

[river elk]

dz = du 1/9

[frigid dust]

im not 100% sure

[river elk]

The problem has 9z^5 9z^2 9z^3

i thought 9z would be easier

then i juts write u^5 u^2 u^3

[frigid dust]

but then you need to eval integral of (u^5 + 4u^2)(u^3 + 1)^12/9 du

and this looks annoying

of course there is always the

binomial bash

method

[river elk]

never heard of that

[frigid dust]

that would be multiplying out this polynomial

dont do that

[frigid dust]

this is better

there is also an even more direct path to the solution

[river elk]

tell me it

[frigid dust]

well, (9z)^3 + 1

[river elk]

would you be able to step by step solve it for me with pen and paper so i could look what you did

i honestly have no clue how to solve it

and i am past the deadlines

no matter what I do or what video i watch i get it wrong and i ma running out of study time cuz i have to work

and if you dont want to to not give me the solution could you do a similar problem or forward me a video where a similar problem is solved

i really cant find anything

sorry to bother

[frigid dust]

so u = ((9z)^3 + 1)

du = 27(9z)^2 dz

the first term has a (9z)^2 in it fortunately

you can write the first term as 1/27 ((9z)^3 + 4) * (27(9z^2))

got it?

i am messing things up

then you can write everything in terms of u and du

[river elk]

@frigid dust

didnt you take ti wrong

no clue how you got du like that

the 27 has to be 3

no?\