Question

Evaluate \(\int \frac{d y}{y(\sqrt{a}-\sqrt{y})},\) for \(a > 0\). (Hint: Use the substitution \(u=\sqrt{y}\) followed by partial fractions.)

Short answer

The indefinite integral of \(\int \frac{d y}{y(\sqrt{a}-\sqrt{y})}\) for \(a > 0\) is \(-\frac{2}{a}\ln\left|\sqrt{y}-1\right| + C\), where C is the constant of integration.

Step-by-step solution

Apply the substitution \(u=\sqrt{y}\)

We are given the hint to use the substitution \(u=\sqrt{y}\), which implies that \(u^2=y\).

Replace \(d y\) with the appropriate expression

To replace \(dy\), we need to find the derivative of \(y\) with respect to \(u\). Since \(u^2=y\), we have: \(2u\,du = dy\) Now, we can rewrite the given integral using the substitution.

Rewrite the integral using \(u\) and find the indefinite integral

Substitute \(u\), and \(2u\,du\) for \(dy\): \(\int \frac{dy}{y(\sqrt{a}-\sqrt{y})} = \int \frac{2u\,du}{u^2(\sqrt{a}-u)}\) Simplify the integral: \(\int \frac{2u\,du}{u^2(\sqrt{a}-u)} = \frac{2}{a} \int \frac{du}{(1-u)}\)

Decompose the integral using partial fractions

Next, we need to perform integration by finding the indefinite integral: \(\frac{2}{a} \int \frac{du}{(1-u)} = -\frac{2}{a} \int \frac{du}{u-1}\)

Integrate each of the terms in partial fractions

Note that the integral is a natural logarithm function: \(-\frac{2}{a} \int \frac{du}{u-1} = -\frac{2}{a}\ln\left|u-1\right| + C\), where \(C\) is the constant of integration.

Replace \(u\) with the original variable \(y\)

Finally, we replace \(u\) with the original variable \(y\) according to the substitution \(u=\sqrt{y}\): \(-\frac{2}{a}\ln\left|u-1\right| + C = -\frac{2}{a}\ln\left|\sqrt{y}-1\right| + C\) The indefinite integral of \(\int \frac{d y}{y(\sqrt{a}-\sqrt{y})}\) for \(a > 0\) is: \(-\frac{2}{a}\ln\left|\sqrt{y}-1\right| + C\)

Key concepts

Substitution Method

The substitution method is a powerful technique in calculus that simplifies integrals by introducing a new variable. In the exercise, we made use of this method by substituting \(u = \sqrt{y}\), which changes the original variable \(y\) to \(u\). This substitution helps us transform the given integral into a simpler form that is easier to evaluate. The basic idea is to change the variable so that the integral becomes one that we can more easily manage.

To effectively use the substitution method:

• Identify a part of the integral whose derivative is present somewhere else in the integral.
• Choose a substitution \(u = g(y)\) that can simplify the expression.
• Differentiate \(u\) to find \(du\), and express \(dy\) in terms of \(du\).
• Substitute these into the integral to rewrite it in terms of \(u\).
• Solve the integral in terms of \(u\), and finally revert back to the original variable \(y\).

This transformation step is crucial as it simplifies the original problem, making it more tractable.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down rational functions into the sum of simpler fractions. This makes integration easier, especially when dealing with fractions that can be expressed as the sum of simpler ones. In our exercise, partial fraction decomposition was the key to solving the integral after the substitution was performed.

Steps to perform partial fraction decomposition:

• Ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial division first.
• Factor the denominator into linear factors if possible.
• Express the fraction as a sum of simpler fractions, where each one has a denominator that's a factor from the previous step.
• Determine the coefficients for these fractions by equating them to the original fraction and solving the resulting system of equations.

With the fractions simplified, integrating each term becomes straightforward, often reducing to basic logarithmic or power functions. This method is especially useful when dealing with irreducible quadratic terms after substitution.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, represent a family of functions whose derivative is the given function. Finding the indefinite integral of a function gives us an expression, plus a constant of integration \(C\), which accounts for all possible antiderivatives.

Key points about indefinite integrals:

• An indefinite integral is represented as \(\int f(x) \, dx = F(x) + C\), where \(F(x)\) is an antiderivative of \(f(x)\).
• The constant \(C\) is crucial because differentiating any constant yields zero, so the antiderivative isn't unique without it.
• Techniques such as substitution and partial fraction decomposition aid in finding these antiderivatives for complex functions.

In the exercise, once our integral was transformed and decomposed, it became straightforward to apply basic integration rules. The result was then expressed in terms of natural logarithms, highlighting the success of combining substitution with partial fractions. Understanding indefinite integrals is fundamental to calculus, providing general solutions to differential problems.