Question

Evaluate the integral
∫
2
x
-
5
x
2
-
5
x
+
6
d
x

Question 5Select one:

I
n
(
2
x
-
5
)
+
c


I
n
(
2
x
2
+
10
x
-
6
)
+
c


I
n
(
2
x
2
+
6
)
+
c


I
n
(
x
2
-
5
x
+
6
)
+
c

Answer 1

To evaluate the integral

\[
\int \frac{2x - 5}{x^2 - 5x + 6} \, dx,
\]

we can start by factoring the denominator \(x^2 - 5x + 6\):

\[
x^2 - 5x + 6 = (x - 2)(x - 3).
\]

Thus, the integral becomes

\[
\int \frac{2x - 5}{(x - 2)(x - 3)} \, dx.
\]

Next, we can use partial fraction decomposition. We express the integrand as

\[
\frac{2x - 5}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3},
\]

for some constants \(A\) and \(B\). Multiplying through by the denominator \((x - 2)(x - 3)\), we have

\[
2x - 5 = A(x - 3) + B(x - 2).
\]

Expanding the right side gives:

\[
2x - 5 = Ax - 3A + Bx - 2B = (A + B)x + (-3A - 2B).
\]

Setting the coefficients equal for \(x\) and the constant terms, we obtain:

1. \(A + B = 2\)
2. \(-3A - 2B = -5\).

Now, we can solve this system of equations. From the first equation, we can express \(B\) in terms of \(A\):

\[
B = 2 - A.
\]

Plugging this into the second equation:

\[
-3A - 2(2 - A) = -5,
\]

which simplifies to:

\[
-3A - 4 + 2A = -5,
\]

leading to:

\[
-A - 4 = -5 \implies -A = -1 \implies A = 1.
\]

Now substituting \(A = 1\) back into the equation for \(B\):

\[
B = 2 - 1 = 1.
\]

So we have \(A = 1\) and \(B = 1\). Therefore, the integrand can be rewritten as:

\[
\frac{2x - 5}{(x - 2)(x - 3)} = \frac{1}{x - 2} + \frac{1}{x - 3}.
\]

Now we can integrate:

\[
\int \left(\frac{1}{x - 2} + \frac{1}{x - 3}\right) \, dx = \int \frac{1}{x - 2} \, dx + \int \frac{1}{x - 3} \, dx = \ln|x - 2| + \ln|x - 3| + C,
\]

where \(C\) is the constant of integration. This can also be combined using properties of logarithms:

\[
= \ln| (x - 2)(x - 3)| + C.
\]

Thus, the final answer is:

\[
\ln| (x - 2)(x - 3)| + C.
\]

Now, let's check which option fits best:

1. \( \ln(2x - 5) + c \)
2. \( \ln(2x^2 + 10x - 6) + c \)
3. \( \ln(2x^2 + 6) + c \)
4. \( \ln(x^2 - 5x + 6) + c \)

Considering the form we derived, the option \( \ln(x^2 - 5x + 6) + c \) can be expressed as \(\ln| (x - 2)(x - 3)| + C\).

Hence, the correct answer is:

\[
\ln(x^2 - 5x + 6) + c.
\]