Question:

Step 1: Decompose the integrand using partial fractions. Let $$ \frac{2x^2+3x+3}{(x+1)(x^2+2x+2)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2} $$ Step 2: Solve for $A$, $B$, $C$. Multiplying both sides by $(x+1)(x^2+2x+2)$ yields: $$ 2x^2+3x+3 = A(x^2+2x+2) + (Bx+C)(x+1) $$ Setting $x=-1$: $$ 2(-1)^2+3(-1)+3 = A((-1)^2+2(-1)+2) + (B(-1)+C)(-1+1) $$ $$ 2-3+3 = A(1-2+2) $$ $$ 2 = A $$ Expanding the right side and comparing coefficients: $$ 2x^2+3x+3 = Ax^2+2Ax+2A + Bx^2+Bx+Cx+C $$ $$ 2x^2+3x+3 = (A+B)x^2 + (2A+B+C)x + (2A+C) $$ Comparing coefficients: For $x^2$: $A+B=2$. Since $A=2$, we have $2+B=2 \Rightarrow B=0$. For $x$: $2A+B+C=3$. Substituting $A=2$ and $B=0$, we get $2(2)+0+C=3 \Rightarrow 4+C=3 \Rightarrow C=-1$. For the constant term: $2A+C=3$. Substituting $A=2$ and $C=-1$, we get $2(2)+(-1)=3 \Rightarrow 4-1=3$, which is consistent. Thus, the partial fraction decomposition is: $$ \frac{2x^2+3x+3}{(x+1)(x^2+2x+2)} = \frac{2}{x+1} + \frac{-1}{x^2+2x+2} $$ Step 3: Write the integral as a sum. $$ Y = \int_0^1 \frac{2}{x+1}\,dx + \int_0^1 \frac{-1}{x^2+2x+2}\,dx $$ $$ Y = 2\int_0^1 \frac{dx}{x+1} - \int_0^1 \frac{dx}{(x^2+2x+1)+1} $$ $$ Y = 2\int_0^1 \frac{dx}{x+1} - \int_0^1 \frac{dx}{(x+1)^2+1} $$ Step 4: Evaluate each integral. The first integral: $$ 2\int_0^1 \frac{dx}{x+1} = 2[\ln|x+1|]_0^1 = 2(\ln|1+1| - \ln|0+1|) = 2(\ln 2 - \ln 1) = 2\ln 2 $$ The second integral: $$ \int_0^1 \frac{dx}{(x+1)^2+1} = [\arctan(x+1)]_0^1 = \arctan(1+1) - \arctan(0+1) = \arctan 2 - \arctan 1 $$ Since $\arctan 1 = \dfrac{\pi}{4}$: $$ \int_0^1 \frac{dx}{(x+1)^2+1} = \arctan 2 - \frac{\pi}{4} $$ Step 5: Combine results. $$ Y = 2\ln 2 - \left(\arctan 2 - \frac{\pi}{4}\right) $$ $$ Y = \frac{\pi}{4} + 2\ln 2 - \arctan 2 $$ Step 6: Express $Y$ in alternative forms. The value of $Y$ is $\dfrac{\pi}{4} + 2\ln 2 - \arctan 2$. We establish the following equivalent forms: 1. Using the identity $\arctan x + \arctan(1/x) = \dfrac{\pi}{2}$ for $x>0$, we have $\arctan 2 = \dfrac{\pi}{2} - \arctan\dfrac{1}{2}$. Substituting this into the expression for $Y$: $$ Y = \frac{\pi}{4} + 2\ln 2 - \left(\frac{\pi}{2} - \arctan\frac{1}{2}\right) $$ $$ Y = \frac{\pi}{4} + 2\ln 2 - \frac{\pi}{2} + \arctan\frac{1}{2} $$ $$ Y = -\frac{\pi}{4} + 2\ln 2 + \arctan\frac{1}{2} $$ Since $\arctan\dfrac{1}{2} = \text{arc cot}\, 2$, we have: $$ Y = -\frac{\pi}{4} + 2\ln 2 + \text{arc cot}\, 2 $$ 2. Using the identity $\arctan A - \arctan B = \arctan\dfrac{A-B}{1+AB}$, we have $\arctan 2 - \arctan 1 = \arctan\dfrac{2-1}{1+2 \cdot 1} = \arctan\dfrac{1}{3}$. Since $\arctan 1 = \dfrac{\pi}{4}$, this implies $\arctan 2 - \dfrac{\pi}{4} = \arctan\dfrac{1}{3}$. Therefore, $\arctan 2 = \arctan\dfrac{1}{3} + \dfrac{\pi}{4}$. Substituting this into the expression for $Y$: $$ Y = \frac{\pi}{4} + 2\ln 2 - \left(\arctan\frac{1}{3} + \frac{\pi}{4}\right) $$ $$ Y = \frac{\pi}{4} + 2\ln 2 - \arctan\frac{1}{3} - \frac{\pi}{4} $$ $$ Y = 2\ln 2 - \arctan\frac{1}{3} $$ Since $\arctan\dfrac{1}{3} = \text{arc cot}\, 3$, we have: $$ Y = 2\ln 2 - \text{arc cot}\, 3 $$ Thus, the integral $Y$ is equal to: $$ \frac{\pi}{4} + 2\ln 2 - \arctan 2 $$ $$ 2\ln 2 - \text{arc cot}\, 3 $$ $$ -\frac{\pi}{4} + 2\ln 2 + \text{arc cot}\, 2 $$