91Ó°ÊÓ

The first step is to decompose the rational function (fraction) given into simpler fractions (partial fractions) in order to make the integration process easier. So we write: $$\frac{20x}{(x-1)(x^2+4x+5)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4x+5}$$ Now we need to determine the constants A, B, and C.

To determine the constants A, B, and C, we multiply both sides of the equation by the common denominator \((x-1)(x^2+4x+5)\): $$20x = A(x^2+4x+5) + (Bx+C)(x-1)$$ Now, we can obtain A, B, and C by substituting different values for x that give us simpler equations to solve for the constants. For example, by setting x=1, we get: $$20 = 5A \Rightarrow A=4$$ Now, we need to find B and C so we can evaluate at some values or compare coefficients: $$20x = 4(x^2+4x+5) + (Bx+C)(x-1)$$ Expanding and comparing coefficients: $$20x = 4x^2+16x+20 + (Bx+C)x - (Bx+C)$$ $$20x = 4x^2+(16+B)x + (20-C)$$ Now, comparing coefficients, we have: $$4=16+B \Rightarrow B=-12$$ $$20=20-C \Rightarrow C=0$$ So, our partial fraction decomposition is: $$\frac{20x}{(x-1)(x^2+4x+5)}=\frac{4}{x-1}+\frac{-12x}{x^2+4x+5}$$

Now, we need to integrate the decomposed integral: $$\int \frac{20 x}{(x-1)\left(x^{2}+4 x+5\right)} d x = \int \frac{4}{x-1}dx -12\int\frac{x}{x^2+4x+5} dx$$ The first integral is a simple natural logarithm, whereas for the second integral, we may use u-substitution. First integral: $$\int \frac{4}{x-1}dx = 4\int\frac{1}{x-1}dx=4ln|x-1|+C_1$$ Now the second integral, there is a u-substitution we can make. Let \(u=x^2+4x+5 \Rightarrow du=(2x+4)dx\). The second integral becomes: $$-12\int\frac{x}{x^2+4x+5} dx=-6\int\frac{2x+4-4}{u}du$$ $$-6\int\frac{2x+4-4}{u}du=-6\int\frac{du}{u}=-6ln|u|+C_2$$ Now, we can substitute back in for u: $$-6ln|x^2+4x+5|+C_2$$

Finally, we can combine the results: $$\int \frac{20 x}{(x-1)\left(x^{2}+4 x+5\right)} d x = 4ln|x-1| - 6ln|x^2+4x+5|+C$$ This is the final solution to the given integral.