Question Statement

Evaluate the integral:

$\int \frac{xdx}{{\left(4x^2+1\right)}^3}$

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Background and Explanation

This problem requires the method of substitution (u-substitution). The key insight is recognizing that the numerator $xdx$ is proportional to the derivative of the inner function $4x^2+1$ appearing in the denominator. When the derivative of the inner function appears as a factor (up to a constant multiple), substitution simplifies the integral to a basic power rule form.

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Solution

Let $I$ denote the given integral:

$I=\int \frac{xdx}{{\left(4x^2+1\right)}^3}$

Step 1: Choose the substitution

We observe that the denominator contains the expression $4x^2+1$, and its derivative is $8x$. Since the numerator contains $xdx$, which differs only by a constant factor, we make the substitution:

$t=4x^2+1$

Step 2: Compute the differential

Differentiating both sides with respect to $x$:

$\frac{dt}{dx}=8x$

Multiplying both sides by $dx$:

$dt=8xdx$

Solving for $xdx$:

$xdx=\frac{1}{8}dt$

Step 3: Substitute into the integral

Replacing $4x^2+1$ with $t$ and $xdx$ with $\frac{1}{8}dt$:

$I=\int \frac{\frac{1}{8}dt}{t^3}$

Factor out the constant $\frac{1}{8}$:

$I=\frac{1}{8}\int \frac{1}{t^3}dt$

Rewrite using negative exponents to apply the power rule:

$I=\frac{1}{8}\int t^{-3}dt$

Step 4: Integrate using the power rule

Applying the power rule for integration $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$ where $n=-3$:

$I=\frac{1}{8}\cdot \frac{t^{-3+1}}{-3+1}+C$

$I=\frac{1}{8}\cdot \frac{t^{-2}}{-2}+C$

Simplify the constants:

$I=\frac{1}{8}\cdot \left(-\frac{1}{2}\right)\cdot t^{-2}+C$

$I=-\frac{1}{16}\cdot \frac{1}{t^2}+C$

Step 5: Substitute back to the original variable

Replace $t$ with $4x^2+1$:

$I=-\frac{1}{16{\left(4x^2+1\right)}^2}+C$

Therefore, the final answer is:

$\boxed{-\frac{1}{16{\left(4x^2+1\right)}^2}+C}$

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Key Formulas or Methods Used

• Substitution Rule: $\int f(g(x))\cdot g^{\prime }(x)dx=\int f(u)du$ where $u=g(x)$
• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne -1$
• Differential Calculation: $d(u)=u^{\prime }(x)dx$

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Summary of Steps

1. Identify the substitution: Let $t=4x^2+1$ because its derivative $8x$ matches the numerator $x$ (up to a constant).
2. Find the differential: Compute $dt=8xdx$, which gives $xdx=\frac{1}{8}dt$.
3. Rewrite the integral: Substitute to get $\frac{1}{8}\int t^{-3}dt$.
4. Integrate: Apply the power rule to obtain $-\frac{1}{16}{t}^{-2}+C$.
5. Back-substitute: Replace $t$ with $4x^2+1$ to get the final answer in terms of $x$.