Question

Evaluate the following integrals. $$\int \frac{x^{2}}{x^{3}-x^{2}+4 x-4} d x$$

Short answer

Question: Evaluate the integral: $$\int \frac{x^{2}}{(x^2+4)(x-1)} d x$$ Answer: $$\int \frac{x^{2}}{(x^2+4)(x-1)} d x = \frac{1}{3}\ln|x-1| + \frac{2}{3}\ln|x^2+4|+ C$$

Step-by-step solution

Factor the denominator

To factor the denominator, look for common factors or use any factoring technique, like factoring by grouping. The denominator is \(x^{3}-x^{2}+4 x-4\). Factoring by grouping, we have: \(x^{3}-x^{2}+4 x-4 = x^2(x-1) + 4(x-1) = (x^2+4)(x-1)\) Thus, the integrand becomes: $$\frac{x^{2}}{(x^2+4)(x-1)}$$

Perform partial fraction decomposition

Now we need to express the fraction as a sum of simpler fractions. To do this we can use the following ansatz: $$\frac{x^{2}}{(x^2+4)(x-1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4}$$ To find the values for A, B, and C, clear the fraction by multiplying through by the denominator and then equating the coefficients of the like terms. \(A(x^2+4) + (Bx+C)(x-1) = x^2\)

Solve for A, B, and C

Multiply and simplify the expression on the left: \(Ax^2 + 4A + Bx^2 - Bx + Cx - C = x^2\) Now, group the coefficients of the like terms: \((A+B)x^2 + (-B+C)x + (4A-C) = x^2\) For the two sides to be equal, the coefficients of similar powers of x must be equal. Thus, we have the following system of equations: $$A+B=1$$ $$-B+C=0$$ $$4A-C=0$$ Solving this system, we get \(A = 1/3\), \(B = 2/3\), and \(C = 2/3\).

Replace A, B, and C in the ansatz

Now, substitute the values of A, B, and C back into the ansatz: $$\frac{x^2}{(x^2+4)(x-1)} = \frac{1/3}{x-1} + \frac{2/3x+2/3}{x^2+4}$$

Integrate each term

Now, integrate each term of the sum separately: $$\int \frac{x^{2}}{(x^2+4)(x-1)} d x = \int \frac{1/3}{x-1} dx + \int \frac{2/3x+2/3}{x^2+4}d x$$ The first term is straightforward to integrate: $$\int \frac{1/3}{x-1} dx = \frac{1}{3}\int \frac{1}{x-1} dx = \frac{1}{3}\ln|x-1|+C_1$$ Now for the second term, we can integrate using a substitution: Let \(u = x^2 + 4\). Then, \(\frac{du}{dx} = 2x\), or \(dx = \frac{du}{2x}\). Now we substitute and integrate: $$\int \frac{2/3x+2/3}{x^2+4}d x = \int \frac{2/3u^{1/2}+2/3}{u}\frac{du}{2u^{1/2}} = \int \frac{2}{3}\frac{1}{u} du = \frac{2}{3} \ln|u| + C_2$$ Substitute back for \(u\): $$\frac{2}{3} \ln|u| + C_2 = \frac{2}{3} \ln|x^2 + 4| + C_2$$

Combine results

Now, combine the results from step 5 to get the final answer: $$\int \frac{x^{2}}{(x^2+4)(x-1)} d x = \frac{1}{3}\ln|x-1| + \frac{2}{3}\ln|x^2+4|+ C$$ Here, \(C = C_1 + C_2\) is the constant of integration.