MAT1581
Assignment 4
2025
, MAT1581 Assignment 4: COMPLETE WORKED
SOLUTIONS
Question 1: Integration Problems (34 marks)
10
1.1 (6 marks): ∫ 𝑑𝑥
25−𝑥 2
Step 1: Factor the denominator 25 − 𝑥 2 = (5) 2 − 𝑥 2 = (5 − 𝑥)(5 + 𝑥)
10 10 𝐴 𝐵
Step 2: Set up partial fraction decomposition = = +
25−𝑥 2 (5−𝑥)(5+𝑥) 5−𝑥 5+𝑥
Step 3: Solve for A and B Multiply both sides by (5 − 𝑥)(5 + 𝑥): 10 = 𝐴(5 + 𝑥) + 𝐵(5 −
𝑥)
Setting 𝑥 = 5 : 10 = 𝐴(10) + 𝐵(0) = 10𝐴 ⇒ 𝐴 = 1 Setting 𝑥 = −5 : 10 = 𝐴(0) + 𝐵(10) =
10𝐵 ⇒ 𝐵 = 1
10 1 1
Step 4: Integrate ∫ 𝑑𝑥 = ∫ 𝑑𝑥 + ∫ 𝑑𝑥
25−𝑥 2 5−𝑥 5+𝑥
1 1 1
For ∫ 𝑑𝑥: Let 𝑢 = 5 − 𝑥 , then 𝑑𝑢 = −𝑑𝑥 ∫ 𝑑𝑥 = ∫ ⋅ (−𝑑𝑢) = −ln|𝑢| + 𝐶 =
5−𝑥 5−𝑥 𝑢
−ln|5 − 𝑥| + 𝐶
1 1
For ∫ 𝑑𝑥: Let 𝑣 = 5 + 𝑥 , then 𝑑𝑣 = 𝑑𝑥 ∫ 𝑑𝑥 = ln|5 + 𝑥| + 𝐶
5+𝑥 5+𝑥
10 5+𝑥
Step 5: Combine results ∫ 𝑑𝑥 = −ln|5 − 𝑥| + ln|5 + 𝑥| + 𝐶 = ln | |+𝐶
25−𝑥 2 5−𝑥
5+𝑥
ANSWER: ln | |+𝐶
5−𝑥
2 −5
1.2 (3 marks): ∫ ln(𝑒 2𝑥 ) 𝑑𝑥
2 −5
Step 1: Apply logarithm property Using the property ln(𝑒 𝑢 ) = 𝑢 : ln(𝑒 2𝑥 ) = 2𝑥 2 − 5
Assignment 4
2025
, MAT1581 Assignment 4: COMPLETE WORKED
SOLUTIONS
Question 1: Integration Problems (34 marks)
10
1.1 (6 marks): ∫ 𝑑𝑥
25−𝑥 2
Step 1: Factor the denominator 25 − 𝑥 2 = (5) 2 − 𝑥 2 = (5 − 𝑥)(5 + 𝑥)
10 10 𝐴 𝐵
Step 2: Set up partial fraction decomposition = = +
25−𝑥 2 (5−𝑥)(5+𝑥) 5−𝑥 5+𝑥
Step 3: Solve for A and B Multiply both sides by (5 − 𝑥)(5 + 𝑥): 10 = 𝐴(5 + 𝑥) + 𝐵(5 −
𝑥)
Setting 𝑥 = 5 : 10 = 𝐴(10) + 𝐵(0) = 10𝐴 ⇒ 𝐴 = 1 Setting 𝑥 = −5 : 10 = 𝐴(0) + 𝐵(10) =
10𝐵 ⇒ 𝐵 = 1
10 1 1
Step 4: Integrate ∫ 𝑑𝑥 = ∫ 𝑑𝑥 + ∫ 𝑑𝑥
25−𝑥 2 5−𝑥 5+𝑥
1 1 1
For ∫ 𝑑𝑥: Let 𝑢 = 5 − 𝑥 , then 𝑑𝑢 = −𝑑𝑥 ∫ 𝑑𝑥 = ∫ ⋅ (−𝑑𝑢) = −ln|𝑢| + 𝐶 =
5−𝑥 5−𝑥 𝑢
−ln|5 − 𝑥| + 𝐶
1 1
For ∫ 𝑑𝑥: Let 𝑣 = 5 + 𝑥 , then 𝑑𝑣 = 𝑑𝑥 ∫ 𝑑𝑥 = ln|5 + 𝑥| + 𝐶
5+𝑥 5+𝑥
10 5+𝑥
Step 5: Combine results ∫ 𝑑𝑥 = −ln|5 − 𝑥| + ln|5 + 𝑥| + 𝐶 = ln | |+𝐶
25−𝑥 2 5−𝑥
5+𝑥
ANSWER: ln | |+𝐶
5−𝑥
2 −5
1.2 (3 marks): ∫ ln(𝑒 2𝑥 ) 𝑑𝑥
2 −5
Step 1: Apply logarithm property Using the property ln(𝑒 𝑢 ) = 𝑢 : ln(𝑒 2𝑥 ) = 2𝑥 2 − 5