91Ó°ÊÓ

To perform partial fraction decomposition, we first write the integral in the form: $$\int \frac{2 x^{2}+5 x+5}{(x+1)\left(x^{2}+2 x+2\right)} d x=\int \frac{A}{x+1}+\frac{Bx+C}{x^{2}+2 x+2} dx$$ To find the constants A, B, and C, we clear the denominators: $$2 x^{2}+5 x+5=A(x^{2}+2 x+2)+(Bx+C)(x+1)$$

Expand the equation and group the terms according to their respective powers of x: $$(2x^2+5x+5)=(Ax^2+2Ax+2A)+((Bx^2+Bx)+(Cx+C))$$ Comparing the coefficients of the powers of x on both sides, we have: $$A+B=2 \space (1)\newline 2A+B+C=5 \space (2)\newline 2A+C=5 \space (3)$$

Using equations (1), (2), and (3), we can solve for A, B, and C. From equation (1), we have: $$A=2-B \space (4)$$ Substitute equation (4) into equation (3): $$2(2-B)+C=5$$ $$4-2B+C=5$$ $$C=2B+1 \space (5)$$ Now substitute equations (4) and (5) into equation (2): $$2(2-B)+2B+1+(2B+1)=5$$ $$5=5$$ Equation (2) is satisfied, meaning the system of equations is correct. We can now solve for A, B, and C using equations (4) and (5). We find that: $$A=1, B=1, C=3$$ Now we can rewrite the integrand using these constants: $$\int \left(\frac{1}{x+1}+\frac{x+3}{x^2+2x+2}\right) dx$$

Now we can integrate the result from step 3 term by term: $$\int \left(\frac{1}{x+1}+\frac{x+3}{x^2+2x+2}\right) dx=\int \frac{1}{x+1} dx + \int \frac{x+3}{x^2+2x+2} dx$$ The first term is a simple natural logarithm: $$\int \frac{1}{x+1} dx=\ln|x+1|+C_1$$ For the second term, we need to complete the square in the denominator: $$\frac{x+3}{x^2+2x+2}=\frac{x+3}{(x+1)^2+1}$$ Let's perform the substitution: \(u=x+1\) $$\int \frac{x+3}{(x+1)^2+1} dx = \int \frac{u+2}{u^2+1}du$$ Now, we can split this one into two simpler fractions and integrate them: $$\int \frac{u+2}{u^2+1}du=\int \frac{u}{u^2+1}du+\int \frac{2}{u^2+1}du$$ For the first integral, we can use the substitution \(v=u^2+1\): $$\int \frac{u}{u^2+1}du=\frac{1}{2}\int \frac{1}{v}dv=\frac{1}{2}\ln|v|+C_2=\frac{1}{2}\ln(u^2+1)+C_2$$ For the second integral, we recognize it as the inverse tangent function: $$\int \frac{2}{u^2+1}du=2\arctan(u)+C_3$$ Putting everything together, we get the final result: $$\int \frac{2 x^{2}+5 x+5}{(x+1)\left(x^{2}+2 x+2\right)} d x = \ln|x+1|+\frac{1}{2}\ln((x+1)^2+1)+2\arctan(x+1)+C$$