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Magazine Chapter 23: Problem 648
Integrate: \(\int\left[\left(8 \mathrm{x}^{3} \mathrm{~d} \mathrm{x}\right) /\left(4 \mathrm{x}^{2}+4 \mathrm{x}+5\right)\right]\)
Short Answer
Expert verified
The integral \(\int\frac{8x^3}{4x^2 + 4x + 5} \mathrm{d}x\) can be evaluated using a substitution method. The final answer is:
\[\frac{1}{4}(\ln|x + 1| - 4\ln|4x^2 + 4x + 5|) + C\]
Step by step solution
01
Choose a substitution
Let's choose the substitution \(u = 4x^2 + 4x + 5\). We want to find an expression for \(\mathrm{d}u\) in terms of \(\mathrm{d}x\).
02
Differentiate the substitution
We now differentiate the substitution expression with respect to \(x\):
\[\frac{d}{dx}(u) = \frac{d}{dx}(4x^2 + 4x + 5)\]
\[\implies \frac{du}{dx} = 8x + 4\]
Now, we want to find an expression for \(\mathrm{d}x\) in terms of \(\mathrm{d}u\):
\[\frac{dx}{du} = \frac{1}{\frac{du}{dx}}\]
03
Calculate \(\mathrm{d}x\) in terms of \(\mathrm{d}u\)
Substitute the expression for \(\frac{du}{dx}\) found in step 2:
\[\frac{dx}{du} = \frac{1}{8x + 4} \implies \mathrm{d}x = \frac{\mathrm{d}u}{8x + 4}\]
04
Perform the substitution
Now substitute the expressions found in steps 1 and 3 into the given integral:
\[\int\frac{8x^3}{u} \cdot \frac{\mathrm{d}u}{8x+4}\]
05
Simplify the integrand
Notice that there is a term \(8x\) common in both numerator and denominator:
\begin{align*}
\int\frac{8x^3}{u} \cdot \frac{\mathrm{d}u}{8x+4} &= \int\frac{x^2}{u} \cdot \frac{8\mathrm{d}u}{8(x+1)} \\
&= \int\frac{x^2}{u} \cdot \frac{\mathrm{d}u}{x+1}
\end{align*}
06
Recognize that \(u\) is a function of \(x\)
Now we realize that the integrand still has an \(x^2\) term, which we can relate back to our substitution using the expression \(u = 4x^2 + 4x + 5\). Solving for \(x^2\), we get:
\begin{align*}
x^2 &= \frac{u - 4x - 5}{4}
\end{align*}
07
Perform the substitution again
We now substitute the expression for \(x^2\) found in step 6 into the integrand:
\begin{align*}
\int\frac{x^2}{u} \cdot \frac{\mathrm{d}u}{x+1} &= \int\frac{\frac{u - 4x - 5}{4}}{u} \cdot \frac{\mathrm{d}u}{x+1} \\
\end{align*}
08
Realize that a different substitution is needed
At this point, we realize that the initial substitution choice was not as helpful as we thought. We need a different substitution to make this integral manageable.
09
Choose a new substitution
Now, let's consider the new substitution \(v = x + 1\). We differentiate:
\[\frac{d}{dx}(v) = \frac{d}{dx}(x + 1) \implies \frac{dv}{dx} = 1\]
We also have,
\[v = x + 1 \implies x = v - 1\]
Now, we will substitute the new substitution into the integral.
10
Perform the new substitution
Now substitute the expressions found in steps 7 and 9 into the integral:
\[\int\frac{\frac{u - 4(v-1) - 5}{4}}{u} \cdot \frac{\mathrm{d}u}{v}\]
11
Simplify the integrand
Now perform the necessary simplifications in the integral:
\begin{align*}
\int\frac{\frac{u - 4(v-1) - 5}{4}}{u} \cdot \frac{\mathrm{d}u}{v} &= \int\frac{u - 4v + 4 - 5}{4uv} ~\mathrm{d}u \\
&= \frac{1}{4} \int \left(\frac{1}{v} - \frac{4}{u}\right) ~\mathrm{d}u
\end{align*}
12
Integrate
Now, we can finally perform the integration:
\begin{align*}
\frac{1}{4} \int \left(\frac{1}{v} - \frac{4}{u}\right) ~\mathrm{d}u &= \frac{1}{4} \left(\int\frac{1}{v} ~\mathrm{d}u - \int\frac{4}{u} ~\mathrm{d}u\right) \\
&= \frac{1}{4}(\ln|v| - 4\ln|u|) + C,
\end{align*}
where C is the constant of integration.
13
Reverse substitutions
Our final task is to replace the substitution expressions back into the result:
\begin{align*}
\frac{1}{4}(\ln|v| - 4\ln|u|) + C &= \frac{1}{4}(\ln|x + 1| - 4\ln|4x^2 + 4x + 5|) + C
\end{align*}
The final answer is:
\[\frac{1}{4}(\ln|x + 1| - 4\ln|4x^2 + 4x + 5|) + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique in calculus used to simplify integrals. It involves changing the variable of integration to help make the integral more manageable. Typically, you replace a complicated expression with a single variable, say \( u \), and solve for \( dx \) in terms of \( du \).
- Choose a substitution: Identify a part of the integral to replace with a single variable \( u \). This is often the hardest step because the right substitution can simplify the problem significantly.
- Differentiate the substitution: Compute \( du/dx \), and then find \( dx \) in terms of \( du \) by rearranging this equation.
- Substitute: Replace the chosen expression and \( dx \) in the integral with \( u \) and \( du \), respectively. Careful adjustment of limits may be required when dealing with definite integrals.
Chain Rule
The chain rule is a fundamental rule in calculus used for differentiating composite functions. It expresses that if a variable \( z \) depends on \( y \), which in turn depends on \( x \), then \( z \) depends on \( x \) through \( y \). The rule is formally expressed as \( \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} \).
- Recognize composite functions: See if one function is nested inside another, such as \( f(g(x)) \).
- Break it down: Determine the derivative of the outer function \( f \) as if it were independent of \( g(x) \), and multiply it by the derivative of the inner function \( g(x) \).
Definite Integral
A definite integral is used to calculate the area under a curve between two limits, \( a \) and \( b \). It results in a numerical value rather than an expression involving an arbitrary constant.
- Set up the integral: Identify the function to integrate and the limits of integration \( a \) and \( b \).
- Evaluate the integral: Compute the antiderivative and then apply the Fundamental Theorem of Calculus.
- Subtract limits: Substitute the upper limit into the antiderivative and subtract the result of substituting the lower limit.
Logarithmic Integration
Logarithmic integration is a technique often encountered when integrating functions of the form \( \frac{1}{x} \). The basic integration of \( \frac{1}{x} \) with respect to \( x \) is \( \ln|x| + C \), where \( C \) is the constant of integration.
- Identify suitable functions: Look for denominators amenable to substitution that, after differentiation, resemble \( \frac{1}{x} \).
- Solve the integral: After substitution, the integral can sometimes simplify to the natural log form, make sure to handle absolute values appropriately due to the properties of logarithms.
- Reverse substitution: Replace the substituted variable back to express the integral in terms of the original variable.
