Problem 22 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d y}{y^{-1}+y^{-3}}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral of the following expression: $$\int \frac{dy}{y^{-1}+y^{-3}}$$ Answer: The integral of the given expression can be evaluated as: $$\int \frac{dy}{y^{-1}+y^{-3}} = \frac{1}{2} \left(\frac{1}{2}(y^2+1)^2 - (y^2+1)\right) + C$$

Step 1: Simplify the Integrands

In this step, we will simplify the expression inside the integral. $$\int \frac{dy}{y^{-1}+y^{-3}} = \int \frac{dy}{\frac{1}{y}+\frac{1}{y^3}} $$ Now, we will combine the terms in the denominator by finding a common denominator: $$\int \frac{dy}{\frac{1}{y}+\frac{1}{y^3}} = \int \frac{dy}{\frac{y^2+1}{y^3}} $$ Now, we can simplify the fraction: $$\int \frac{dy}{\frac{y^2+1}{y^3}} = \int y^3dy$$ Step 2: Perform a substitution

Step 2: Substitution

We perform the substitution $$u=y^2+1$$Then, we find the derivative of u with respect to y: $$\frac{du}{dy} = 2y$$Now, let's rewrite our integral in terms of u: $$\int y^3dy = \int \frac{y^2}{2} \frac{du}{dy}dy$$Substitute u for \(y^2\) in the integral:$$\int \frac{y^2}{2} \frac{du}{dy}dy = \int \frac{u-1}{2} \frac{du}{dy}dy$$Now, substitute du for \(\frac{du}{dy}dy\):$$\int \frac{u-1}{2} \frac{du}{dy}dy = \int \frac{u-1}{2} du$$ Step 3: Integrate with respect to u

Step 3: Integration

Now we can proceed to integrate the expression with respect to u: $$\int \frac{u-1}{2} du =\frac{1}{2}\int (u-1) du $$Now, we will integrate term by term:$$\frac{1}{2}\int (u-1) du = \frac{1}{2} \left(\int u \ du - \int 1 \ du\right)$$After integrating both terms, we get:$$\frac{1}{2} \left(\int u \ du - \int 1 \ du\right) = \frac{1}{2} \left(\frac{1}{2}u^2 - u\right) + C$$where C is the constant of integration. Step 4: Convert back to y

Step 4: Back Substitution

Finally, we will convert our result back into terms of y using our substitution:$$\frac{1}{2} \left(\frac{1}{2}u^2 - u\right) + C = \frac{1}{2} \left(\frac{1}{2}(y^2+1)^2 - (y^2+1)\right) + C$$This is the final result of our integral: $$\int \frac{dy}{y^{-1}+y^{-3}} = \frac{1}{2} \left(\frac{1}{2}(y^2+1)^2 - (y^2+1)\right) + C$$