Question Statement

Evaluate the integral:

$\int \frac{dx}{(1-3x{)}^2}$

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Background and Explanation

This integral involves a composite function where the inner function is linear $(1-3x)$. The substitution method (u-substitution) is the appropriate technique here, as it allows us to transform the integral into a simpler power rule form. Recognizing that the derivative of the inner function $(-3)$ is a constant makes this substitution particularly straightforward.

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Solution

Let $I$ denote the integral we wish to evaluate:

$I=\int \frac{dx}{(1-3x{)}^2}$

To simplify this integral, we use the substitution method. Notice that the denominator contains the expression $(1-3x)$, whose derivative is the constant $-3$. This makes it an ideal candidate for substitution.

Step 1: Choose the substitution

Let $t=1-3x$

Step 2: Compute the differential

Differentiating both sides with respect to $x$: $\frac{dt}{dx}=-3$

Using differentials, we obtain: $dt=-3dx$

Solving for $dx$: $dx=-\frac{1}{3}dt$

Step 3: Substitute into the integral

Replacing $(1-3x)$ with $t$ and $dx$ with $-\frac{1}{3}dt$ in equation (1):

$\begin{array}{rl}I& =\int \frac{1}{t^2}\cdot \left(-\frac{1}{3}\right)dt\\ & =-\frac{1}{3}\int t^{-2}dt\end{array}$

Step 4: Apply the power rule for integration

Using the power rule $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$ for $n\ne -1$, with $n=-2$:

$\begin{array}{rl}I& =-\frac{1}{3}\cdot \frac{t^{-2+1}}{-2+1}+c\\ & =-\frac{1}{3}\cdot \frac{t^{-1}}{-1}+c\\ & =-\frac{1}{3}\cdot \left(-\frac{1}{t}\right)+c\\ & =\frac{1}{3}\cdot \frac{1}{t}+c\end{array}$

Step 5: Substitute back to the original variable

Recall that $t=1-3x$. Substituting back:

$I=\frac{1}{3(1-3x)}+c$

Therefore, the final solution is:

$\int \frac{dx}{(1-3x{)}^2}=\frac{1}{3(1-3x)}+c$

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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 of Linear Function: $d(ax+b)=adx$

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Summary of Steps

1. Identify substitution: Let $t=1-3x$ to simplify the composite expression in the denominator
2. Compute differential: Differentiate to get $dt=-3dx$, then solve for $dx=-\frac{1}{3}dt$
3. Rewrite integral: Substitute to obtain $I=-\frac{1}{3}\int t^{-2}dt$
4. Integrate: Apply the power rule to get $-\frac{1}{3}\cdot \frac{t^{-1}}{-1}+c=\frac{1}{3t}+c$
5. Back-substitute: Replace $t$ with $(1-3x)$ to obtain the final answer $\frac{1}{3(1-3x)}+c$