The integral $\int \frac { x - 3 } { x ^ { 2 } - 6 x + 10 } d x$ is

A) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 4 } + C$
B) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 2 } + C$
C) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 3 } + C$
D) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 5 } + C$
E) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 6 } + C$

Question

The integral $\int \frac { x - 3 } { x ^ { 2 } - 6 x + 10 } d x$ is

A) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 4 } + C$
B) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 2 } + C$
C) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 3 } + C$
D) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 5 } + C$
E) $\frac { \ln \left| x ^ { 2 } - 6 x + 10 \right| } { 6 } + C$

Quizplus · Accepted Answer

The integral can be solved by completing the square in the denominator and recognizing the derivative of the resulting expression in the numerator. Completing the square for $x^2 - 6x + 10$ gives $(x-3)^2 + 1$. The derivative of $x^2 - 6x + 10$ is $2x - 6$, which matches the numerator $x - 3$ up to a constant factor. Therefore, the integral is of the form $\frac{1}{2}\ln|x^2 - 6x + 10| + C$, which simplifies to choice B.

The Integral ∫x−3x2−6x+10dx\int \frac { x - 3 } { x ^ { 2 } - 6 x + 10 } d x∫x2−6x+10x−3​dx

The Integral $\int \frac { x - 3 } { x ^ { 2 } - 6 x + 10 } d x$