Question

Refer to the summary box (Partial Fraction Decompositions) and evaluate the following integrals. $$\int \frac{x}{(x-1)\left(x^{2}+2 x+2\right)^{2}} d x$$

Short answer

$$\int \frac{x}{x^3 + 2x^4+x^5} dx$$ Short Answer: The integral of the given function is: $$-\frac{1}{2}\ln\left| 1-\frac{1}{x} \right| +\frac{1}{4}\arctan(x+1) + \frac{1}{8} \frac{1}{x+1} + C$$

Step-by-step solution

Divide the numerator and denominator

Divide the numerator and denominator by x, so that the integrand takes the following form: $$\int \frac{1}{(\frac{x}{x}-\frac{1}{x})\left(\frac{x^2}{x}+2 \frac{x^2}{x}+2\right)^{2}} d x$$

Perform the partial fraction decomposition

Let's perform the partial fraction decomposition. The given integral becomes: $$\int \frac{1}{(1-\frac{1}{x})(x^2+2x+2)^2} dx$$ Now, we can write it as: $$\int \frac{A}{1-\frac{1}{x}} + \frac{Bx+C}{(x^2+2x+2)}+\frac{Dx+E}{(x^2+2x+2)^2} dx$$ where we need to find A, B, C, D, and E.

Solve for A, B, C, D, and E

To find the constants A, B, C, D, and E, follow these steps: 1. Clear fractions by multiplying both sides by \((x^2+2x+2)^2(1-\frac{1}{x})\). 2. Divide the resulting polynomial equation into different parts. 3. Find A, B, C, D, and E by equating the coefficients of the corresponding powers of x. After doing this, we get: $$A=-\frac{1}{2}, \qquad B=0, \qquad C=\frac{1}{4}, \qquad D=0, \qquad E=-\frac{1}{4}$$

Rewrite the integral with constants

Now that we have the constants, let's rewrite the given integral with these constants: $$\int -\frac{1}{2}\frac{dx}{1-\frac{1}{x}} +\frac{1}{4}\frac{dx}{(x^2+2x+2)} -\frac{1}{4}\frac{dx}{(x^2+2x+2)^2}$$

Integrate the simpler fractions

Let's integrate each simpler fraction individually: 1. \(\int -\frac{1}{2}\frac{dx}{1-\frac{1}{x}}\): Substitute \(u=1-\frac{1}{x}\), then integrate. 2. \(\int \frac{1}{4}\frac{dx}{(x^2+2x+2)}\): Complete the square in the denominator, then use the substitution \(u=x+1\). 3. \(\int -\frac{1}{4}\frac{dx}{(x^2+2x+2)^2}\): Complete the square in the denominator and use the substitution \(t=x+1\). After integrating the fractions, we get: $$-\frac{1}{2}\ln\left| 1-\frac{1}{x} \right| +\frac{1}{4}\arctan(x+1) + \frac{1}{8} \frac{1}{x+1} + C$$ Now, combining all the fractions, we get our final answer: $$-\frac{1}{2}\ln\left| 1-\frac{1}{x} \right| +\frac{1}{4}\arctan(x+1) + \frac{1}{8} \frac{1}{x+1} + C$$