Find $\int x \sqrt[5]{3x+2} dx$ using substitution (help me understand this particular way)

Question

$$\int x \sqrt[5]{3x+2} dx$$

I'm reading a book that solves this by making $t=\sqrt[5]{3x+2}$. Then it does:

$$t^5 = 3x+2\\ 5t^4 dt = 3dx$$

The rest is just isolating dx, then replacing it in th expression, then taking $t^5 = 3x+2$ and doing the same for x. * What I don't get is the step above. The book says that what was done was that the derivative was taken on both sides and that this was justified by the chain rule (I didn't get what they meant).

I played around with it and I think I figured out why, can someone tell me if I am correct? I know this notation can be treated similarly to fractions (or so i read) so here's what I did:

$$t^5 = 3x+2 \Leftrightarrow t^5\frac{1}{dt}*\frac{1}{dx} = (3x+2)*\frac{1}{dt}*\frac{1}{dx} \Leftrightarrow 5t^4 *\frac{1}{dx}=3*\frac{1}{dt} \Leftrightarrow 5t^4dt = 3dx$$

In the last step you multiply both sides by dt and dx to get rid of the fractions.

Is this correct? If yes can someone explain to me exactly why this works?

Answer 1

It's maybe easier this way : $$t^5=3x+2 \implies x=\frac 1 3(t^5-2)$$ Differentiate now $$\frac {dx}{dt}=\frac d {dt} \{\frac 1 3(t^5-2)\}$$ $$\frac {dx}{dt}=\frac 5 3t^4$$ $$3dx= 5 t^4 dt$$

Answer 2

More or less we can think of it as $$\frac{d}{dt}(f(t))=\frac{d}{dx}\cdot \frac{dx}{dt}(f(t))$$

So we get $$\frac{d}{dx}(t^5)=\frac{d}{dx}(3x+5)$$

$$\frac{d}{dt}(t^5)\frac{dt}{dx}=\frac{d}{dx}(3x+5)$$

$$5t^4\frac{dt}{dx}=3$$

$$5t^4dt=3dx$$

Also note the approach of change of variable where you let $u=3x+2$ and swap to a rational function. Then note that $du=3dx$ by the same question you were asking. But $x=\frac{u-2}{3}$ and the integral becomes $$\int x\sqrt[5]{3x+2}dx=\int \frac{u-2}{3}\sqrt[5]{u}(\frac{du}{3})=\frac{1}{9}\int (u-2)u^{1/5}du$$ which is quite easy to deal with (granted I made no arithmetic mistakes -- and I think this approach makes it easier to avoid them though the idea is the same).

Answer 3

What you did is correct, but it's really unnecessary. With the book's substitution, $\color{purple}{t = \sqrt[5]{3x+2}}$. Once you have $t^5 = 3x+2$, you want to differentiate both sides with respect to $t$:

$$\frac{\mathrm d}{\mathrm dx} t^5 = \frac{\mathrm d}{\mathrm dx} (3x+2) \iff 5t^4 \cdot \frac{\mathrm dt}{\mathrm dx} = 3 \iff \color{blue}{\frac{5}{3}t^4 \mathrm dt = \mathrm dx}$$

The second step is simply using the Chain Rule: $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm dy}{\mathrm dt} \cdot \dfrac{\mathrm dt}{\mathrm dx}$.

From here, $\color{green}{x = \dfrac{t^5-2}{3}}$, so with the substitution, the integral can be rewritten:

$$\int \color{green}{x}\color{purple}{\sqrt[5]{3x+2}}\color{blue}{\mathrm dx} \to \int \color{green}{\frac{t^5-2}{3}}\cdot\color{purple}{t}\cdot\color{blue}{\frac{5}{3}t^4 \mathrm dt} = \int \frac{5\left(t^{10}-2t^5\right)}{9} \mathrm dt$$

And this just leaves evaluating the simpler indefinite integral above and then plugging $x$ back in.