1. ## antiderivative help

hey guys, a little confused on the order of this one. any step by step help is appreciated.

find the antiderivative:
$\displaystyle f(x) = (x^2 + 1)\sqrt[4]{x^3 + 3x + 2}$

2. Originally Posted by JGaraffa
hey guys, a little confused on the order of this one. any step by step help is appreciated.

find the antiderivative:
$\displaystyle f(x) = (x^2 + 1)\sqrt[4]{x^3 + 3x + 2}$
Use the substitution rule: let $\displaystyle z := x^3+3x+2$. It follows that $\displaystyle dz = 3x^2+3=3(x^2+1)\,dx$, thus you have that

$\displaystyle \int (x^2 + 1)\sqrt[4]{x^3 + 3x + 2}\, dx = \tfrac{1}{3}\cdot \int \sqrt[4]{x^3 + 3x + 2}\cdot 3 (x^2 + 1)\, dx =\tfrac{1}{3}\cdot \int \sqrt[4]{z}\, dz=\ldots$

3. Thanks!!

4. Originally Posted by JGaraffa
hey guys, a little confused on the order of this one. any step by step help is appreciated.

find the antiderivative:
$\displaystyle f(x) = (x^2 + 1)\sqrt[4]{x^3 + 3x + 2}$
Observe that $\displaystyle 3(x^2+1)$ is the derivative of $\displaystyle x^3+3x+2$, so:

$\displaystyle \frac{d}{dx}(x^3+3x+2)^{5/4}=3(x^2+1)\times \frac{5}{4}\times (x^3+3x+2)^{1/4}$

etc...

CB