Problem 7 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d x}{(3-5 x)^{4}}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral of the function $$\frac{1}{(3-5x)^4}$$ with respect to x. Answer: $$\int \frac{dx}{(3-5 x)^{4}}= \frac{1}{15(3-5x)^3} + C$$

Step 1: Assign a substitution

Let's substitute the expression inside the integral to make it easier to solve. Define a variable u as follows: $$u=3-5x \Rightarrow x = \frac{1}{5}(3-u)$$ Next, we need to find the derivative of x with respect to u: $$\frac{dx}{du}=-\frac{1}{5}$$ Thus, the differential dx is given by: $$dx=-\frac{1}{5}du$$

Step 2: Rewrite the integral using substitution

Now, we substitute the expressions found in step 1 into the integral: $$\int \frac{1}{(3-5x)^4}dx = \int \frac{1}{u^4}\left(-\frac{1}{5}\right)du$$ Simplify the integral: $$\int \frac{1}{u^4}\left(-\frac{1}{5}\right)du = -\frac{1}{5}\int \frac{1}{u^4} du$$

Step 3: Integrate the simplified expression

Now, integrate the expression: $$-\frac{1}{5}\int \frac{1}{u^4} du =-\frac{1}{5}\int u^{-4} du$$ The integral of $$u^{-4}$$ is given by $$-\frac{1}{3u^3} + C$$, where C is the constant of integration. So, the integral on the left side becomes: $$-\frac{1}{5}\left(-\frac{1}{3u^3}\right) + C$$

Step 4: Substitute back to find the integral in terms of x

We know that u = 3 - 5x, so substitute this back into the expression: $$-\frac{1}{5}\left(-\frac{1}{3(3-5x)^3}\right) + C$$ Simplify the expression: $$\int \frac{1}{(3-5x)^4} dx = \frac{1}{15(3-5x)^3} + C$$ Thus, the evaluated integral is: $$\int \frac{dx}{(3-5 x)^{4}}= \frac{1}{15(3-5x)^3} + C$$