# Thread: Evaluating this indefinite integral

1. ## Evaluating this indefinite integral

It's the indefinite integral of (3x-7)^1.7 dx I don't even know where I'd begin finding that antiderivative, maybe substitution of some sort? Could someone walk me through this?

2. Originally Posted by fattydq
It's the indefinite integral of (3x-7)^1.7 dx I don't even know where I'd begin finding that antiderivative, maybe substitution of some sort? Could someone walk me through this?
Let $\displaystyle u=3x-7$ and use the power rule. This one is pretty easy.

3. Originally Posted by fattydq
It's the indefinite integral of (3x-7)^1.7 dx I don't even know where I'd begin finding that antiderivative, maybe substitution of some sort? Could someone walk me through this?
$\displaystyle \int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}$

Let $\displaystyle u = 3x - 7$ so that $\displaystyle \frac{du}{dx} = 3$.

So the integral becomes

$\displaystyle \frac{1}{3}\int{u^{1.7}\frac{du}{dx}\,dx} = \frac{1}{3}\int{u^{1.7}\,du}$.

I trust you can go from here

4. Originally Posted by Reckoner
Let $\displaystyle u=3x-7$ and use the power rule. This one is pretty easy.
So it would end up being 1/3 the integral of u^1.7 which would end up being (u^2.7)/2.7 times 1/3, with u resub'd back in? so 1/3 times (3x-7)^2.7 over 2.7?

Originally Posted by Prove It
$\displaystyle \int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}$

Let $\displaystyle u = 3x - 7$ so that $\displaystyle \frac{du}{dx} = 3$.

So the integral becomes

$\displaystyle \frac{1}{3}\int{u^{1.7}\frac{du}{dx}\,dx} = \frac{1}{3}\int{u^{1.7}\,du}$.

I trust you can go from here
So my final answer would just be 1/3 times (3x-7)^2.7/2.7?

5. Originally Posted by fattydq
So my final answer would just be 1/3 times (3x-7)^2.7/2.7?
Yes, plus an arbitrary constant.