Evaluating this indefinite integral

**fattydq** · February 25th 2009, 03:23 PM

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?

**Reckoner** · February 25th 2009, 03:28 PM

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 $u=3x-7$ and use the power rule. This one is pretty easy.

**Prove It** · February 25th 2009, 03:35 PM

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?

$\int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}$

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

So the integral becomes

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

I trust you can go from here

**fattydq** · February 25th 2009, 03:36 PM

Originally Posted by Reckoner

Let $u=3x-7$ 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

$\int{(3x - 7)^{1.7}\,dx} = \frac{1}{3}\int{(3x - 7)^{1.7}\cdot 3\,dx}$

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

So the integral becomes

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

I trust you can go from here

So my final answer would just be 1/3 times (3x-7)^2.7/2.7?

**Reckoner** · February 25th 2009, 04:01 PM

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.

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