Question

Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{1} \frac{y \ln ^{4}\left(y^{2}+1\right)}{y^{2}+1} d y$$

Short answer

Short Answer: The integral $\int_{0}^{1} \frac{y \ln ^{4}\left(y^{2}+1\right)}{y^{2}+1} d y$ can be evaluated in terms of another integral $I = 4 \int_0^1 y \ln^3(y^2 +1) \arctan(y) dy$, which cannot be expressed explicitly using elementary functions. Thus, the final answer for the original integral is given by $\frac{\pi \ln^4(2)}{4} - 4I$. Since the integrand is positive, absolute values are not needed in the final answer.

Step-by-step solution

Use integration by parts

In order to solve the integral, we will first use integration by parts. Let's choose: \(u = \ln^4(y^2 + 1)\), then \(du = \frac{4 \ln^3(y^2 + 1)}{y^2 + 1} \cdot 2y \, dy\) \(v = \frac{1}{y^2+1} dy\), then \(v = \arctan(y)\) (using the known antiderivative). Using integration by parts (\(\int u dv = uv - \int v du\)), we have $$\int_{0}^{1} \frac{y \ln^4 (y^2 + 1)}{(y^2 + 1)} dy = \left[\ln ^4(y^2 +1)\arctan(y)\right]_{0}^{1} - 4 \int_0^1 y \ln^3(y^2 +1) \arctan(y) dy$$.

Evaluate the first part of the integral

Now, we need to evaluate the first part: $$\left\lbrack \ln^4(y^2 +1) \arctan(y) \right\rbrack_{0}^{1} = \left(\ln^4(2) \arctan(1) - \ln^4 (1) \arctan(0)\right) = \left(\ln^4(2) \cdot \frac{\pi}{4} - 0\right) = \frac{\pi \ln^4(2)}{4}$$

Simplify and evaluate the remaining integral

Let's name the remaining integral \(I\): $$I = 4 \int_0^1 y \ln^3(y^2 +1) \arctan(y) dy$$ Unfortunately, we cannot evaluate this integral explicitly using elementary functions. However, the problem asked for an answer that includes absolute values only when needed. Since we started with a positive integrand, there will be no need for absolute values in the final answer.

Combine the results and write the final answer

Now, we can combine the results from Steps 2 and 3. Given that the integral \(I\) is not elementary, we can only express the final answer in terms of I: $$\int_{0}^{1} \frac{y \ln ^{4}\left(y^{2}+1\right)}{y^{2}+1} d y = \frac{\pi \ln^4(2)}{4} - 4I$$

Key concepts

Definite Integrals

Definite integrals are a type of integral in calculus that evaluate the area under a curve over a specified interval. In simpler terms, they provide the exact value of the integral from a starting point to an ending point, which are called the lower and upper limits of the integral. Definite integrals are denoted by the notation \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits.

Some important features of definite integrals include:

• The result of a definite integral is a real number, not a function.
• They can be used to calculate areas, volumes, and other physical properties that accumulate over a range.
• In the Fundamental Theorem of Calculus, definite integrals relate the rate of change (derivative) of a function to the function itself.

In the given problem, the definite integral is taken over the interval from 0 to 1, and it involves a complex expression containing logarithmic and trigonometric components. The evaluation of this definite integral showcases integration techniques, including integration by parts, to find the exact area under the curve defined by the function \( \frac{y \ln^4(y^2 + 1)}{y^2 + 1} \).

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. They are critical in various areas of mathematics and applied sciences. The logarithmic function \( \ln(x) \) refers to the natural logarithm, which is the logarithm with base \( e \). In calculus, dealing with logarithmic functions often requires careful manipulation and use of rules of logarithms.

Key points about logarithmic functions include:

• They can handle a wide range of numbers, providing a way to simplify multiplication by turning it into addition.
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \), which is frequently used in integration.
• When raising a logarithm to a power, as seen in \( \ln^4(y^2 + 1) \), it involves the power rule and product rule in differentiation or integration processes.

In the integral problem, we work with \( \ln^4(y^2 + 1) \), demonstrating how to handle powers of logarithmic terms using integration by parts. This involves differentiating \( \ln^4(y^2 + 1) \) and handling terms with precision to ensure correct evaluation. Although evaluating such integrals can be challenging, they highlight the versatility and complexity of calculus.

Trigonometric Functions

Trigonometric functions are mathematical functions of angles commonly used in calculus and related disciplines. They include sine, cosine, and tangent, among others. These functions are periodic and are fundamental in modeling oscillations and waves.

Here are some important properties:

• Tangent (\( \tan(x) \)) is one of the primary trigonometric functions and is defined as \( \frac{\sin(x)}{\cos(x)} \).
• The integral of \( \tan(x) \) leads to the well-known antiderivative \( \arctan(x) \), commonly encountered in integration problems.
• Trigonometric functions have useful identities that simplify integration and differentiation tasks.

In this exercise, trigonometric functions come into play when finding the antiderivative with integration by parts. Specifically, the transformation \( \int \frac{1}{y^2 + 1} \, dy = \arctan(y) \) is used to ease integration. These trigonometric solutions demonstrate how calculus integrates different areas like geometry and algebra, showing the interconnectedness of mathematical concepts.