Find the integral. Select the correct answer.
$\int \frac { x ^ { 2 } - x + 2 } { x ^ { 3 } - 2 x ^ { 2 } + x } d x$
A) $\ln \left| \frac { x ^ { 2 } } { x - 1 } \right| - \frac { 2 } { x - 1 } + C$
B) $\ln \left| \frac { x ^ { 2 } } { x - 1 } \right| - \frac { 1 } { x - 1 } + C$
C) $\ln \left| \frac { x - 1 } { x ^ { 2 } } \right| - \frac { 1 } { x - 1 } + C$
D) $\ln \left| \frac { x - 1 } { x ^ { 2 } } \right| - \frac { 2 } { x - 1 } + C$

Question

Quizplus · Accepted Answer

Explanation:Factor the denominator: $x^3 - 2x^2 + x = x(x^2 - 2x + 1) = x(x - 1)^2$. Then use partial fractions to decompose the integrand:$$\frac{x^2 - x + 2}{x(x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}$$Multiplying both sides by $x(x - 1)^2$ gives$$x^2 - x + 2 = A(x - 1)^2 + Bx(x - 1) + C x$$Substituting $x = 0$ gives $2 = -A$, so $A = -2$.Substituting $x = 1$ gives $1 = C$.Equating coefficients of $x^2$ gives $1 = A + B$, so $B = 3$.Therefore,$$\int \frac{x^2 - x + 2}{x(x - 1)^2} dx = \int \left(-\frac{2}{x} + \frac{3}{x - 1} + \frac{1}{(x - 1)^2}\right) dx$$$$= -2\ln|x| + 3\ln|x - 1| - \frac{1}{x - 1} + C$$$$= \ln\left|\frac{(x - 1)^3}{x^2}\right| - \frac{1}{x - 1} + C$$$$= \ln\left|\frac{x^2}{x - 1}\right| - \frac{2}{x - 1} + C$$

Find the Integral ∫x2−x+2x3−2x2+xdx\int \frac { x ^ { 2 } - x + 2 } { x ^ { 3 } - 2 x ^ { 2 } + x } d x∫x3−2x2+xx2−x+2​dx

Find the Integral $\int \frac { x ^ { 2 } - x + 2 } { x ^ { 3 } - 2 x ^ { 2 } + x } d x$