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x32x2+1(x22x+5)2dx\int\frac{x^{3}-2x^{2}+1}{(x^{2}-2x+5)^{2}}dx

Full solution

Q. x32x2+1(x22x+5)2dx\int\frac{x^{3}-2x^{2}+1}{(x^{2}-2x+5)^{2}}dx
  1. Given Integral: We are given the integral: \newlinex32x2+1(x22x+5)2dx\int\frac{x^3 - 2x^2 + 1}{(x^2 - 2x + 5)^2}dx\newlineTo solve this integral, we will use the method of partial fractions. However, the denominator (x22x+5)2(x^2 - 2x + 5)^2 does not factor into linear terms over the real numbers because the discriminant 224152^2 - 4\cdot1\cdot5 is negative. Therefore, we cannot decompose the fraction into simpler partial fractions directly. Instead, we will complete the square for the denominator and then try to simplify the integral.
  2. Complete the Square: First, let's complete the square for the denominator:\newlinex22x+5=(x22x+1)+4=(x1)2+4x^2 - 2x + 5 = (x^2 - 2x + 1) + 4 = (x - 1)^2 + 4\newlineNow, the integral becomes:\newline(x32x2+1)dx((x1)2+4)2\int\frac{(x^3 - 2x^2 + 1)dx}{((x - 1)^2 + 4)^2}
  3. Substitution: Next, we will make a substitution to simplify the integral. Let: u=x1u = x - 1 Then, du=dxdu = dx, and x=u+1x = u + 1 Now we can rewrite the integral in terms of uu: ((u+1)32(u+1)2+1)du/(u2+4)2\int((u + 1)^3 - 2(u + 1)^2 + 1)du / (u^2 + 4)^2
  4. Expand and Simplify: We expand the numerator to simplify the expression:\newline(u3+3u2+3u+1)2(u2+2u+1)+1(u2+4)2du\int\frac{(u^3 + 3u^2 + 3u + 1) - 2(u^2 + 2u + 1) + 1}{(u^2 + 4)^2}\,du\newline= (u3+3u2+3u+12u24u2+1)du(u2+4)2\int\frac{(u^3 + 3u^2 + 3u + 1 - 2u^2 - 4u - 2 + 1)\,du}{(u^2 + 4)^2}\newline= (u3+u2u)du(u2+4)2\int\frac{(u^3 + u^2 - u)\,du}{(u^2 + 4)^2}
  5. Split into Integrals: Now, we will split the integral into three separate integrals: u3du(u2+4)2\int \frac{u^3 \, du}{(u^2 + 4)^2} + u2du(u2+4)2\int \frac{u^2 \, du}{(u^2 + 4)^2} - udu(u2+4)2\int \frac{u \, du}{(u^2 + 4)^2}
  6. Complex Integral: The first integral (u3du)/(u2+4)2\int(u^3 \, du) / (u^2 + 4)^2 is a bit complex and does not have a straightforward antiderivative. We will need to use a method such as integration by parts or a further substitution. However, this is a non-standard integral and may require a special function or a numerical method to solve. At this point, we realize that the approach we have taken may not be the most efficient, and we may need to reconsider our strategy.