Problem 50
Question
Find the indefinite integral. $$\int x^{3}\left(x^{2}+1\right)^{3 / 2} d x$$
Step-by-Step Solution
Verified Answer
The short answer for the indefinite integral of \(\int x^{3}(x^{2}+1)^{\frac{3}{2}} dx\) is: \(\frac{1}{7}(x^2+1)^{\frac{7}{2}} - \frac{1}{5}(x^2+1)^{\frac{5}{2}} + C\).
1Step 1: Choose a substitution
Let's choose the substitution \(u = x^2 + 1\). This will simplify the expression inside the power and help us to perform the integral.
2Step 2: Find the differential
To find the differential, take the derivative of \(u\) with respect to \(x\) and multiply by \(dx\):
\[ du = \frac{d}{dx}(x^2+1) dx = (2x) dx\]
Now, solve for \(dx\):
\[ dx = \frac{du}{2x} \]
3Step 3: Rewrite the integral in terms of u and du
Now, substitute \(u\) and \(du\) in the integral and simplify:
\[
\int x^3 (x^2+1)^{\frac{3}{2}} dx = \int x^3 u^{\frac{3}{2}} \cdot \frac{du}{2x}
\]
The \(x^3\) and the \(2x\) cancel out, resulting in:
\[ \int \frac{1}{2}x^2 u^{\frac{3}{2}} du \]
Since \(u = x^2 + 1\), we can rewrite \(x^2\) in terms of u:
\[ x^2 = u - 1\]
Now, substitute for \(x^2\):
\[ \int \frac{1}{2}(u - 1)u^{\frac{3}{2}} du \]
4Step 4: Integrate with respect to u
Now, we can integrate the expression with respect to \(u\):
\[ \frac{1}{2} \int (u - 1)u^{\frac{3}{2}} du = \frac{1}{2} \int (u^{\frac{5}{2}} - u^{\frac{3}{2}}) du \]
Integrate term by term:
\[
\frac{1}{2} \left[ \int u^{\frac{5}{2}} du - \int u^{\frac{3}{2}} du \right] = \frac{1}{2} \left[\frac{2}{7}u^{\frac{7}{2}} - \frac{2}{5}u^{\frac{5}{2}}\right] + C
\]
5Step 5: Replace u with original expression
Now, substitute the original expression of \(u\) back into the result:
\[
\frac{1}{2} \left[ \frac{2}{7}(x^2+1)^{\frac{7}{2}} - \frac{2}{5}(x^2+1)^{\frac{5}{2}} \right] + C\]
Finally, simplify the expression:
\[
\frac{1}{7}(x^2+1)^{\frac{7}{2}} - \frac{1}{5}(x^2+1)^{\frac{5}{2}} + C
\]
The indefinite integral of \(x^3(x^2+1)^{\frac{3}{2}}\) is:
\[
\int x^{3}(x^{2}+1)^{\frac{3}{2}} dx = \frac{1}{7}(x^2+1)^{\frac{7}{2}} - \frac{1}{5}(x^2+1)^{\frac{5}{2}} + C
\]
Key Concepts
Substitution MethodIntegration TechniquesCalculus
Substitution Method
The substitution method is a key technique in calculus for simplifying the process of integration. By substituting one expression with another, integration can become more manageable and less complex. Let's take a closer look at how it works.
Substitution involves replacing a part of the integral with a new variable, often simplifying the integral into a standard form that is easier to solve. For example, when we set \(u = x^2 + 1\), we change the variable of integration from \(x\) to \(u\). This can make expounded equations much simpler. The chosen substitution should ideally make most of the initial variables disappear or simplify drastically, paving the way for a straightforward integral.
Substitution involves replacing a part of the integral with a new variable, often simplifying the integral into a standard form that is easier to solve. For example, when we set \(u = x^2 + 1\), we change the variable of integration from \(x\) to \(u\). This can make expounded equations much simpler. The chosen substitution should ideally make most of the initial variables disappear or simplify drastically, paving the way for a straightforward integral.
When to Use the Substitution Method?
- When an integral features a function and its derivative.
- When compound expressions can be simplified into basic algebraic forms.
- When trigonometric identities or logarithmic identities seem applicable.
Integration Techniques
Integration, the inverse of differentiation, involves finding a function whose derivative is the given function. To tackle a wide variety of problems in calculus, several integration techniques are often employed.
Common Integration Techniques
- Substitution: Simplifies integrals by introducing a new variable.
- Integration by Parts: Useful when dealing with the product of two functions, applying the formula \( \int u\, dv = uv - \int v\, du \).
- Partial Fractions: Breaks down rational functions into simpler fractions that can be integrated individually.
- Trigonometric Integrals: Deals with integrals of functions involving trigonometric functions, often using identities.
Calculus
Calculus, founded on the concepts of differentiation and integration, underpins much of modern mathematics. It provides tools to handle continuous change, enabling the calculation of areas and volumes, the analysis of functions, and much more.
In calculus, integration is the process of finding integrals, which are the reverse operations of differentiation. This operation provides a way to accumulate quantities, be it areas under curves or solutions to differential equations.
In calculus, integration is the process of finding integrals, which are the reverse operations of differentiation. This operation provides a way to accumulate quantities, be it areas under curves or solutions to differential equations.
Importance of Calculus
- It facilitates the understanding of dynamic systems in physics, economics, biology, and engineering.
- It allows the solving of real-world problems involving rates of change and areas under curves.
- Calculus is essential in the creation of models for scientific phenomena.
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