Integral of x^2(3x^3+4)^4
and (x^4)/(3x^5+2)
Thanks! =)
here is the first, the second one is done similarly. as for the way we plug in our substitution, some people do it differently. this is the way i like to do it. tell me if you get it
$\displaystyle \int x^2 (3x^3 + 4)^4~dx$
Let $\displaystyle u = 3x^3 + 4$
$\displaystyle \Rightarrow du = 9x^2~dx$
$\displaystyle \Rightarrow \frac 19~du = x^2~dx$
So our integral becomes:
$\displaystyle \frac 19 \int u^4~du$
so as you see. i wrote the 1/9 du instead of x^2 dx, i then replaced what was in the brackets by u, as my substitution demands.
$\displaystyle \frac 19 \int u^4~du = \frac 19 \cdot \frac 15u^5 + C$
$\displaystyle = \frac 1{45}u^5 + C$
Back-substitute:
$\displaystyle = \frac 1{45}(3x^3 + 4)^5 + C$
what you did is what u-substitution imitates. it is more or less the formalization of your method. some students cannot see that however, so substitution is best for beginners i think. but yes, you are correct, that is a valid way to do the problem, and a way Krizalid may have suggested
I attained this the same way through constant manipulation $\displaystyle \int\frac{x^4}{3x^5+2}dx=\frac{1}{15}\int\frac{15x ^4}{3x^5+2}dx\Rightarrow{\frac{1}{15}ln|3x^5+2|+C}$ I was able to just integrate because once again I had an integral of the form $\displaystyle \int{f(g(x))\cdot{g'(x)}dx}$