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Problem 12

# Evaluate the integrals. $$\int(x+1)\left[\cos \left(x^{2}+2 x\right)+\left(x^{2}+2 x\right)\right] d x$$

Expert verified
The short answer for the integral $$\int(x+1)\left[\cos \left(x^{2}+2 x\right)+\left(x^{2}+2 x\right)\right] d x$$ is: $$\frac{1}{2}\sin{x^2+2x} + \frac{1}{4}x^4 + x^3 + x^2 + C$$
See the step by step solution

## Step 1: Separate the integral

We have: $$\int(x+1)\left[\cos \left(x^{2}+2 x\right)+\left(x^{2}+2 x\right)\right] d x = \int(x+1)\cos \left(x^{2}+2 x\right) dx + \int(x+1)\left(x^{2}+2 x\right) dx$$

## Step 2: Evaluate the integral of the first part

Now, to find the integral of the first part: $$\int(x+1)\cos \left(x^{2}+2 x\right) dx$$ Here, we have to use the substitution method: Let $$u = x^2 + 2x$$, then $$\frac{d u}{d x} = 2x + 2$$. Hence, the integral becomes $$\int \frac{x+1}{2x+2}\cos{(u)} d u$$ Let's simplify: $$\int \frac{x+1}{2x+2}\cos{(u)} d u = \frac{1}{2}\int\cos{(u)} d u$$ Now, integrate the expression: $$\frac{1}{2}\int\cos{(u)} d u = \frac{1}{2}\sin{(u)} + C_1$$ Let's substitute back the value of u: $$\frac{1}{2}\sin{x^2+2x} + C_1$$

## Step 3: Evaluate the integral of the second part

Now, to find the integral of the second part: $$\int(x+1)\left(x^{2}+2 x\right) dx$$ Integration using polynomial rules: $$\int(x+1)\left(x^{2}+2 x\right) dx = \int x^3 + 3x^2 + 2x dx$$ Now we integrate each term: $$\frac{1}{4}x^4 + x^3 + x^2 + C_2$$

## Step 4: Combine the results

Now, we combine the results of the two integrals: $\frac{1}{2}\sin{x^2+2x} + \frac{1}{4}x^4 + x^3 + x^2 + C$ where $$C = C_1 + C_2$$ is the constant of integration. So, the final result is: $$\int(x+1)\left[\cos \left(x^{2}+2 x\right)+\left(x^{2}+2 x\right)\right] d x = \frac{1}{2}\sin{x^2+2x} + \frac{1}{4}x^4 + x^3 + x^2 + C$$

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