# Integral question

#### evan4888

Here is the problem:

$$\int ^2_1 \frac{dx}{(3-5x)^2}$$

$$u = 3-5x$$

$$du = -5dx$$

$$dx = -\frac{1}{5} du$$

so, (with the 7 and 2 being negatives)

= $$-\frac{1}{5} \int^7_2 \frac{du}{u^2}$$

= $$-\frac{1}{5} (-\frac{1}{u})]^7_2$$

But what I don't understand is how the $$u^2$$ becomes just $$u$$.

#### Jameson

Why did you change your bounds? You had everything right.

$$\frac{1}{u^2}$$ can be written as $$u^{-2}$$

Now follow the general power rule and $$\int u^{-2}du = \frac{(u^{-1})}{(-1)}$$

But to write that more nicely write it as $$-\frac{1}{u}$$

Make sense?

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