integration by substitution-messy problem!

[crzymath]

problem asks to integrate by substitution
this is what i have:

?x[sup:2svmhf2k]3[/sup:2svmhf2k](x[sup:2svmhf2k]4[/sup:2svmhf2k]+3)[sup:2svmhf2k]2[/sup:2svmhf2k]dx

u = x[sup:2svmhf2k]4[/sup:2svmhf2k] + 3
du/dx = 4x --> du = 4xdx

now i substitue it back in:
= ?x[sup:2svmhf2k]3[/sup:2svmhf2k](u[sup:2svmhf2k]2[/sup:2svmhf2k]) du
= 1/4 ? 3x[sup:2svmhf2k]2[/sup:2svmhf2k]/2 (2u[sup:2svmhf2k]3[/sup:2svmhf2k]/3) dx

this is where i get confused because i think i messed it up since the answer is suppose to be: (x[sup:2svmhf2k]4[/sup:2svmhf2k]+ 3)[sup:2svmhf2k]3[/sup:2svmhf2k]/12 + C

any clue where i went wrong?

 

[Deleted member 4993]

problem asks to integrate by substitution
this is what i have:

?x[sup:3gmmdfgg]3[/sup:3gmmdfgg](x[sup:3gmmdfgg]4[/sup:3gmmdfgg]+3)[sup:3gmmdfgg]2[/sup:3gmmdfgg]dx

u = x[sup:3gmmdfgg]4[/sup:3gmmdfgg] + 3
du/dx = 4x --> du = 4xdx<<<< Incorrect

du/dx = 4 * x[sup:3gmmdfgg]3[/sup:3gmmdfgg]

Now it should be simpler


now i substitue it back in:
= ?x[sup:3gmmdfgg]3[/sup:3gmmdfgg](u[sup:3gmmdfgg]2[/sup:3gmmdfgg]) du
= 1/4 ? 3x[sup:3gmmdfgg]2[/sup:3gmmdfgg]/2 (2u[sup:3gmmdfgg]3[/sup:3gmmdfgg]/3) dx

this is where i get confused because i think i messed it up since the answer is suppose to be: (x[sup:3gmmdfgg]4[/sup:3gmmdfgg]+ 3)[sup:3gmmdfgg]3[/sup:3gmmdfgg]/12 + C

any clue where i went wrong?

 

[soroban]

Hello, crzymath!

I'll baby-step through it for you . . .

$\displaystyle \text{Integrate by substitution: }\;\int x^3(x^4+3)^2\,dx$

$\displaystyle \text{Let: }\:u \:=\:x^4+3 \quad\Rightarrow\quad du \:=\:4x^3\,dx \quad\Rightarrow\quad x^3\,dx \:=\:\frac{du}{4}$

$\displaystyle \text{We have: }\;\;\int\underbrace{(x^4+3)^2}_{\downarrow}\underbrace{(x^3\,dx)}_{\downarrow}$

$\displaystyle \text{Substitute: }\;\;\int \;\;u^2\quad\;\;\left(\frac{du}{4}\right) \;\;=\;\;\tfrac{1}{4}\int u^2\,du\;\;=\;\;\tfrac{1}{12} u^3 + C$


$\displaystyle \text{Back-substitute: }\;\tfrac{1}{12}(x^4+3)^3 + C$