Solve Integral: \int^2_1 \frac{dx}{(3-5x)^2}

[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?

 

[M98Ranger]

In this integral, the substitution u = 3-5x is used to simplify the integral. By substituting u for 3-5x, the integral becomes:

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

Using the substitution u = 3-5x, the denominator (3-5x)^2 becomes u^2. This is because when we substitute u for 3-5x, we are essentially replacing the entire expression of 3-5x with u, including the exponent of 2. Therefore, the u^2 term is just a simplified version of (3-5x)^2.

I hope this explanation helps to clarify the use of u in this integral.