Note: Example 3.29 as well as "the General Method" from the textbook
Links to an external site.
might help you with this question.

Problem:

Evaluate the integral ∫
x+4

(x+2)(x−4)
dx

A student hands in the following 'solution'. Is it correct?

Line 1: Let's use the method of Partial Fractions. The degree of the numerator is less than the degree of the denominator, so we do not need to use long division.

Line 2: There are two distinct (=different) linear factors so the partial fraction decomposition is of the form:

Line 3:
x+4

(x+2)(x−4)
=
A

x+2
+
B

x−4
for some constants A and B.

Line 4: Multiply both sides by (x+2)(x−4):

Line 5: x+4=A(x−4)+B(x+2)

Line 6: Simplify: x+4=Ax−4A+Bx+2B

Line 7: The number of x on the two sides must be equal. Also, the constants on both sides must be equal:

  A+B=1

−4A+2B=4

Line 8: The first equation becomes B=1−A which can be inserted into the second equation:

Line 9: −4A+2(1−A)=4⟹−4A+2−2A=4⟹−6A=2⟹A=−
1

3

Line 10: Therefore B=1−(−
1

3
)=
4

3

Line 11: Now we know the values of A and B so we can continue with the integral:

Line 12: ∫
x+4

(x+2)(x−4)
dx=∫(
4

3(x+2)
−
1

3(x−4)
)dx=

Line 13: =∫
4

3(x+2)
dx−∫
1

3(x−4)
dx=

Line 14: =
4

3
ln|x+2|−
1

3
ln|x−4|+C

Is the solution correct, or if not, in which line does the first error occur? [line]

Question

Note: Example 3.29 as well as "the General Method" from the textbook
Links to an external site.
 might help you with this question.

Problem:

Evaluate the integral ∫
x+4

(x+2)(x−4)
dx

A student hands in the following 'solution'. Is it correct?

Line 1: Let's use the method of Partial Fractions. The degree of the numerator is less than the degree of the denominator, so we do not need to use long division.

Line 2: There are two distinct (=different) linear factors so the partial fraction decomposition is of the form:

Line 3: 
x+4

(x+2)(x−4)
=
A

x+2
+
B

x−4
    for some constants A  and B.

Line 4: Multiply both sides by (x+2)(x−4):

Line 5: x+4=A(x−4)+B(x+2)

Line 6: Simplify: x+4=Ax−4A+Bx+2B

Line 7: The number of x on the two sides must be equal. Also, the constants on both sides must be equal:

A+B=1

−4A+2B=4

Line 8: The first equation becomes B=1−A  which can be inserted into the second equation:

Line 9: −4A+2(1−A)=4⟹−4A+2−2A=4⟹−6A=2⟹A=−
1

3

Line 10: Therefore B=1−(−
1

3
)=
4

3

Line 11: Now we know the values of A and B so we can continue with the integral:

Line 12: ∫
x+4

(x+2)(x−4)
dx=∫(
4

3(x+2)
−
1

3(x−4)
)dx=

Line 13: =∫
4

3(x+2)
dx−∫
1

3(x−4)
dx=

Line 14: =
4

3
ln|x+2|−
1

3
ln|x−4|+C

Is the solution correct, or if not, in which line does the first error occur? [line]

BlackTom AI · Accepted Answer

The question asks us to evaluate the student’s partial fractions solution and identify where the first error occurs.
First, restating the problem: We need to integrate ∫ (x+4)/[(x+2)(x−4)] dx. The student proposes a partial fraction decomposition of the form (x+4)/[(x+2)(x−4)] = A/(x+2) + B/(x−4), then proceeds to solve for A and B.
Analyzing each step:
Line 1: The student states we will use partial fractions since the degree of the numerator (1) is less than the degree of the denominator (2). This is correct.
Line 2: With two distinct linear factors, the form A/(x+2) + B/(x−4) is appropriate. This setup is correct.
Line 3: The decomposition is written as (x+4)/[(x+2)(x−4)] = A/(x+2) + B/(x−4). This is the standard correct setup.
Line 4–6: Multiplying both sides by the denominator and collecting terms yields x+4 = A(x−4) + B(x+2). Expanding gives x+4 = Ax − 4A + Bx + 2B. This is correct algebra.
Line 7: Equating coefficients of like powers of x yields two equations: A + B = 1 and −4A + 2B = 4. This is the proper system to solve for A and B. So far, everything is logically sound.
Line 8–10: Solving A + B = 1 and −4A + 2B = 4 leads to B = 1 − A, and substituting gives −4A + 2(1 − A) = 4 → −6A + 2 = 4 → −6A = 2 → A = −1/3. Then B = 1 − (−1/3) = 4/3. These arithmetic steps are correct; A and B are found correctly.
Line 11: The student claims to proceed with the integral using these A and B values. This is the natural next step, but the form in which the integral is written matters.
Line 12: Here is where the first error appears. The student writes the integral as ∫ (4/3)(x+2) − (1/3)(x−4) dx. This expression incorrectly treats the partial fraction terms as if A/(x+2) and B/(x−4) were proportional to (x+2) and (x−4) themselves, rather than as constants divided by (x+2) and (x−4). The correct integrand should be A/(x+2) + B/(x−4) = (−1/3)/(x+2) + (4/3)/(x−4).
So, line 12 is the first line where the student’s representation deviates from the correct form; the error is due to miswriting the decomposition from constants over linear factors to a linear function over those factors inside the integral.
In summary, the error sits at line 12, where the integrand should be written as (−1/3)·1/(x+2) + (4/3)·1/(x−4) instead of (4/3)(x+2) − (1/3)(x−4). This misrepresentation leads to an incorrect antiderivative, since integrating 1/(x+2) or 1/(x−4) yields logarithms, not linear terms in (x+2) or (x−4).

MAT136H5 S 2025 - All Sections 3.4 Preparation Check

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