Question

Preliminary steps The following integrals require a preliminary step such as a change of variables before using the method of partial fractions. Evaluate these integrals. $$\int \frac{d y}{y(\sqrt{a}-\sqrt{y})}, a>0(\text {Hint: Let } u=\sqrt{y}.)$$

Short answer

The solution to the integral is $$2\sqrt{y} + 2\ln|a-\sqrt{y}| + C$$.

Step-by-step solution

Change of Variables

We are given the hint to use the change of variables \(u = \sqrt{y}\). To do this, we will first find \(du\) in terms of \(dy\). Differentiate \(u = \sqrt{y}\) with respect to \(y\): \(du/dy = \frac{1}{2\sqrt{y}}\) Now, we can write \(dy\) in terms of \(du\): \(dy = 2\sqrt{y}du = 2u du\) Now, substitute the change of variables \(u = \sqrt{y}\) and \(dy = 2u du\) into the integral: $$\int \frac{d y}{y(\sqrt{a}-\sqrt{y})} = \int \frac{2udu}{u^2(a-u)}$$

Perform Partial Fraction Decomposition

We will now perform the partial fraction decomposition on the integrand: $$\frac{2u}{u^2(a - u)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{a-u}$$ Multiplying both sides by \(u^2(a-u)\), we get: $$2u = Au(a-u) + Bu(a-u) + Cu^2$$ Now we need to find the constants \(A, B,\) and \(C\). We can do this by equating coefficients of same powers of \(u\): 1. For \(u^2\)-terms: \(C = 2\) 2. For \(u^1\)-terms: \(A+B=-2\) 3. For \(u^0\)-terms: \(Aa = 0\) From (3), it is clear that \(A = 0\). Now, from (2), we can figure out that \(B = -2\). Therefore, our partial fraction decomposition becomes: $$\frac{2u}{u^2(a - u)} = -\frac{2}{u^2} + \frac{2}{a-u}$$

Evaluate the Integral

Now we can rewrite the integral: $$\int \frac{2u}{u^2(a - u)} du = -2\int \frac{1}{u^2} du + 2\int \frac{1}{a-u} du$$ Perform the integration: $$= -2\left(-\frac{1}{u}\right) + 2\ln|a-u| + C$$ Now we will go back to the original variable \(y\). Remember, \(u = \sqrt{y}\), so \(y = u^2\), which implies \(1/u = \sqrt{y}\), and \(a-u = a-\sqrt{y}\): $$= 2\sqrt{y} + 2\ln|a-\sqrt{y}| + C$$ So, the solution to the integral is: $$\int \frac{d y}{y(\sqrt{a}-\sqrt{y})} = 2\sqrt{y} + 2\ln|a-\sqrt{y}| + C$$

Key concepts

Change of Variables

The change of variables is a helpful technique in calculus to simplify the evaluation of integrals. It involves substituting a new variable for an existing one in order to transform the integral into a simpler form. In our problem, the hint given is to let \( u = \sqrt{y} \). This particular substitution is aimed at tackling the square root present in the integral.

This makes the process easier by allowing us to express both \( y \) and its differential \( dy \) in terms of the new variable \( u \). Here's how it works:
- Differentiate \( u = \sqrt{y} \) with respect to \( y \) to find the relationship between \( du \) and \( dy \).
- This gives us \( du/dy = \frac{1}{2\sqrt{y}} \), which leads to \( dy = 2u \, du \).

Substituting these into the integral transforms it into a function of \( u \) rather than \( y \), simplifying the process and often making it ready for further techniques, such as partial fractions, to be applied.

Method of Partial Fractions

The method of partial fractions is a technique in calculus used to break down complex rational expressions into simpler fractions, making them easier to integrate. After performing a change of variables, we often end up with an integrable expression that requires simplification, particularly when polynomial denominators are involved.

This method is especially useful when dealing with rational expressions, which are fractions where the numerator and the denominator are both polynomials. The process generally involves:

• Expressing the rational function as a sum of simpler fractions.
• Solving for unknown constants by equating coefficients.
• Decomposing the original expression so that it becomes a sum of these simpler fractions.

In our example, after the change of variables, the expression \( \frac{2u}{u^2(a-u)} \) is decomposed using partial fractions. This allows us to express it as:
\( \frac{A}{u} + \frac{B}{u^2} + \frac{C}{a-u} \).

Determining these coefficients involves multiplying through by the common denominators and equating coefficients, which helps isolate each term and solve the integral step by step.

Partial Fraction Decomposition

Partial fraction decomposition is the process of breaking down a complex fraction into a sum or difference of simpler fractions. When the method of partial fractions is applied, decomposition provides a way to simplify the integrands, making the integration of each term feasible with basic techniques.

The steps usually include:

• Assuming the form of the partial fraction expansion.
• Equating the original fraction to this expansion.
• Finding values for the constants that make the equality true by equating coefficients.

This rigorous yet straightforward approach transforms a complicated fraction like \( \frac{2u}{u^2(a-u)} \) into simpler terms that are easier to integrate:
\(-\frac{2}{u^2} + \frac{2}{a-u} \).

From here, you can integrate each term individually. In this case, each term corresponds to basic integral formulas, namely \( \int \frac{1}{u^2} du \) and \( \int \frac{1}{a-u} du \). Each term is straightforward to integrate, resulting in a vastly simplified approach to solving the original integral.