Question Statement

Evaluate the integral:

$\int \left(3x^2+2x\right){\left(x^3+x^2+9\right)}^{5}dx$

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Background and Explanation

This integral requires recognizing the pattern where the integrand consists of a function raised to a power multiplied by its own derivative. This is a direct application of the reverse chain rule (substitution method) for integration.

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Solution

Let $I=\int \left(3x^2+2x\right){\left(x^3+x^2+9\right)}^{5}dx$.

First, observe the structure of the integrand. Notice that if we let $f(x)=x^3+x^2+9$, then its derivative is: $f^{\prime }(x)=\frac{d}{dx}(x^3+x^2+9)=3x^2+2x$

This matches the first factor in our integrand. Therefore, we can rewrite the integral as: $I=\int {\left(x^3+x^2+9\right)}^5\cdot \left(3x^2+2x\right)dx=\int [f(x){]}^5\cdot f^{\prime }(x)dx$

Using the integration formula for this specific form (which is essentially $u$-substitution where $u=x^3+x^2+9$): $\int [f(x){]}^n\cdot f^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}+c$

Here, $n=5$, so $n+1=6$. Applying the formula: $I=\frac{{\left(x^3+x^2+9\right)}^{5+1}}{5+1}+c$

Simplifying the expression: $I=\frac{{\left(x^3+x^2+9\right)}^6}{6}+c$

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Key Formulas or Methods Used

• Reverse Chain Rule / Substitution Method: $\int f^n(x)\cdot f^{\prime }(x)dx=\frac{f^{n+1}(x)}{n+1}+c$
• Power Rule for Integration: Applied after substitution
• Recognition Pattern: Identifying that $3x^2+2x$ is the derivative of $x^3+x^2+9$

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Summary of Steps

1. Identify the inner function: Let $u=x^3+x^2+9$ and verify that $du=(3x^2+2x)dx$ matches the remaining factor.
2. Rewrite the integral: Express as $\int u^{5}du$.
3. Apply the power rule: Integrate to get $\frac{u^6}{6}+c$.
4. Substitute back: Replace $u$ with $x^3+x^2+9$ to obtain the final answer $\frac{(x^3+x^2+9{)}^6}{6}+c$.

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Question 17

Question Statement

Evaluate the integral:

$\int \left(5e^{5x}-x^{-3}+3^{2x}\right)dx$

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Background and Explanation

This problem combines three basic integration techniques: exponential functions with linear arguments, power rule for negative exponents, and exponential functions with base $a$ requiring substitution. Each term must be integrated separately using the appropriate rule.

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Solution

Let $I=\int \left(5e^{5x}-x^{-3}+3^{2x}\right)dx$.

We can split this into three separate integrals: $I=5\int e^{5x}dx-\int x^{-3}dx+\int 3^{2x}dx$

Step 1: Integrate the first term

For $\int e^{5x}dx$, using the rule $\int e^{ax}dx=\frac{e^{ax}}{a}$: $5\int e^{5x}dx=5\cdot \frac{e^{5x}}{5}=e^{5x}$

Step 2: Integrate the second term

For $\int x^{-3}dx$, using the power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ where $n=-3$: $-\int x^{-3}dx=-\frac{x^{-3+1}}{-3+1}=-\frac{x^{-2}}{-2}=\frac{x^{-2}}{2}=\frac{1}{2x^2}$

Step 3: Integrate the third term

For $\int 3^{2x}dx$, we use substitution: Let $t=2x$, then $dt=2dx$ or $dx=\frac{dt}{2}$.

Substituting: $\int 3^{2x}dx=\int 3^t\cdot \frac{dt}{2}=\frac{1}{2}\int 3^{t}dt$

Using the formula $\int a^{x}dx=\frac{a^x}{\ln a}$: $\frac{1}{2}\int 3^{t}dt=\frac{1}{2}\cdot \frac{3^t}{\ln 3}+c=\frac{3^{2x}}{2\ln 3}+c$

Step 4: Combine all terms

Putting everything together: $I=e^{5x}+\frac{1}{2x^2}+\frac{3^{2x}}{2\ln 3}+c$

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Key Formulas or Methods Used

• Exponential Integration: $\int e^{ax}dx=\frac{e^{ax}}{a}+c$
• Power Rule: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ (for $n\ne -1$)
• General Exponential Rule: $\int a^{x}dx=\frac{a^x}{\ln a}+c$
• Substitution Method: Used for $3^{2x}$ by letting $t=2x$ to linearize the exponent

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Summary of Steps

1. Separate the integral: Break into three parts: $5e^{5x}$, $-x^{-3}$, and $3^{2x}$.
2. First term: Integrate $e^{5x}$ to get $\frac{e^{5x}}{5}$, multiply by 5 to get $e^{5x}$.
3. Second term: Apply power rule to $x^{-3}$ to get $\frac{x^{-2}}{-2}$, simplify signs to get $+\frac{1}{2x^2}$.
4. Third term: Substitute $t=2x$, integrate $3^t$ to get $\frac{3^t}{\ln 3}$, substitute back and include the factor of $\frac{1}{2}$ to get $\frac{3^{2x}}{2\ln 3}$.
5. Combine: Add all results and include the constant of integration $+c$.