Free. Exclusive. Just for you.
Four unique services that make learning easier, faster, and smarter - only on our website.

Integral Calculus Worksheets :: 25 - 48 - Free Printable

Integral Calculus Worksheets :: 25 - 48

Educational worksheet: Integral Calculus Worksheets :: 25 - 48. Download and print for classroom or home learning activities.

PNG 300×225 46.4 KB Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #1352621
Show Answer Key & Explanations Step-by-step solution for: Integral Calculus Worksheets :: 25 - 48
Let’s solve each integral one by one. We’ll go step by step, carefully checking our work.

---

1. ∫ (x² + 2)² dx

First, expand the square:

(x² + 2)² = x⁴ + 4x² + 4

Now integrate term by term:

∫ x⁴ dx = x⁵/5
∫ 4x² dx = 4·x³/3 = (4/3)x³
∫ 4 dx = 4x

So answer: x⁵/5 + (4/3)x³ + 4x + C

---

2. ∫ (x³ - x + 5)/√x dx

Rewrite √x as x^(1/2), so divide each term:

= ∫ [x³/x^(1/2) - x/x^(1/2) + 5/x^(1/2)] dx
= ∫ [x^(5/2) - x^(1/2) + 5x^(-1/2)] dx

Integrate:

∫ x^(5/2) dx = x^(7/2)/(7/2) = (2/7)x^(7/2)
∫ x^(1/2) dx = x^(3/2)/(3/2) = (2/3)x^(3/2) → but with minus sign: - (2/3)x^(3/2)
∫ 5x^(-1/2) dx = 5 · x^(1/2)/(1/2) = 10x^(1/2)

Answer: (2/7)x^(7/2) - (2/3)x^(3/2) + 10√x + C

---

3. ∫ (x² + 2)³ / x dx

Expand numerator: (x² + 2)³ = x⁶ + 6x⁴ + 12x² + 8

Divide by x: x⁵ + 6x³ + 12x + 8/x

Integrate:

∫ x dx = x⁶/6
∫ 6x³ dx = 6·x⁴/4 = (3/2)x⁴
∫ 12x dx = 6x²
∫ 8/x dx = 8 ln|x|

Answer: x⁶/6 + (3/2)x⁴ + 6x² + 8 ln|x| + C

---

4. ∫ (x² - 1)/(x + 1) dx

Note: x² - 1 = (x - 1)(x + 1), so cancel (x + 1):

= ∫ (x - 1) dx = x²/2 - x + C

Answer: x²/2 - x + C

---

5. ∫ (sin x + cos x) dx

Simple: ∫ sin x dx = -cos x, ∫ cos x dx = sin x

Answer: -cos x + sin x + C

---

6. ∫ (eˣ + e⁻ˣ) dx

∫ eˣ dx = eˣ, ∫ e⁻ˣ dx = -e⁻ˣ

Answer: eˣ - e⁻ˣ + C

---

7. ∫ (1/(1+x²)) dx

Standard integral: arctan(x)

Answer: arctan(x) + C

---

8. ∫ (sec²x + csc²x) dx

∫ sec²x dx = tan x, ∫ csc²x dx = -cot x

Answer: tan x - cot x + C

---

9. ∫ (x² + 1)/(x² - 1) dx

Do polynomial division or rewrite:

(x² + 1)/(x² - 1) = [ (x² - 1) + 2 ] / (x² - 1) = 1 + 2/(x² - 1)

Now, 2/(x² - 1) = 2/[(x-1)(x+1)] → use partial fractions:

2/[(x-1)(x+1)] = A/(x-1) + B/(x+1)

Multiply both sides: 2 = A(x+1) + B(x-1)

Set x = 1: 2 = A(2) → A = 1
Set x = -1: 2 = B(-2) → B = -1

So: 1 + 1/(x-1) - 1/(x+1)

Integrate: ∫1 dx = x, ∫1/(x-1) dx = ln|x-1|, ∫-1/(x+1) dx = -ln|x+1|

Answer: x + ln|x-1| - ln|x+1| + C or x + ln|(x-1)/(x+1)| + C

---

10. ∫ (x³ - 2x + 1)/(x² + 1) dx

Polynomial division:

Divide x³ - 2x + 1 by x² + 1.

x³ ÷ x² = x → multiply: x(x² + 1) = x³ + x
Subtract: (x³ - 2x + 1) - (x³ + x) = -3x + 1

So: x + (-3x + 1)/(x² + 1)

Break into: x - 3x/(x² + 1) + 1/(x² + 1)

Integrate:

∫x dx = x²/2
∫ -3x/(x² + 1) dx → let u = x² + 1, du = 2x dx → so -3/2 ∫ du/u = -3/2 ln|u| = -3/2 ln(x² + 1)
∫ 1/(x² + 1) dx = arctan(x)

Answer: x²/2 - (3/2) ln(x² + 1) + arctan(x) + C

---

11. ∫ (x⁴ - 1)/(x² + 1) dx

Factor numerator: x⁴ - 1 = (x² - 1)(x² + 1)

Cancel (x² + 1): left with x² - 1

∫ (x² - 1) dx = x³/3 - x + C

Answer: x³/3 - x + C

---

12. ∫ (x² + 2x + 1)/(x + 1) dx

Numerator is (x + 1)², denominator is (x + 1) → simplifies to (x + 1)

∫ (x + 1) dx = x²/2 + x + C

Answer: x²/2 + x + C

---

13. ∫ (x³ + 8)/(x + 2) dx

Note: x³ + 8 = (x + 2)(x² - 2x + 4) → sum of cubes

Cancel (x + 2): left with x² - 2x + 4

∫ (x² - 2x + 4) dx = x³/3 - x² + 4x + C

Answer: x³/3 - x² + 4x + C

---

14. ∫ (x² - 4)/(x - 2) dx

x² - 4 = (x - 2)(x + 2) → cancel (x - 2) → left with x + 2

∫ (x + 2) dx = x²/2 + 2x + C

Answer: x²/2 + 2x + C

---

15. ∫ (x⁴ - 16)/(x² - 4) dx

x⁴ - 16 = (x² - 4)(x² + 4) → cancel (x² - 4) → left with x² + 4

∫ (x² + 4) dx = x³/3 + 4x + C

Answer: x³/3 + 4x + C

---

16. ∫ (x³ - 27)/(x - 3) dx

x³ - 27 = (x - 3)(x² + 3x + 9) → difference of cubes

Cancel (x - 3): left with x² + 3x + 9

∫ (x² + 3x + 9) dx = x³/3 + (3/2)x² + 9x + C

Answer: x³/3 + (3/2)x² + 9x + C

---

17. ∫ (x² + 4x + 4)/(x + 2) dx

Numerator: (x + 2)² → cancel one (x + 2) → left with x + 2

∫ (x + 2) dx = x²/2 + 2x + C

Answer: x²/2 + 2x + C

---

18. ∫ (x³ + 1)/(x + 1) dx

x³ + 1 = (x + 1)(x² - x + 1) → sum of cubes

Cancel (x + 1): left with x² - x + 1

∫ (x² - x + 1) dx = x³/3 - x²/2 + x + C

Answer: x³/3 - x²/2 + x + C

---

19. ∫ (x⁴ - 1)/(x - 1) dx

x⁴ - 1 = (x - 1)(x³ + x² + x + 1) → factor using difference of squares twice? Or just divide.

Actually: x⁴ - 1 = (x² - 1)(x² + 1) = (x - 1)(x + 1)(x² + 1)

So dividing by (x - 1) leaves (x + 1)(x² + 1) = x³ + x² + x + 1

Wait — better to do polynomial division or recognize pattern.

Alternatively: since x⁴ - 1 divided by x - 1 is geometric series: 1 + x + x² + x³

Yes! Because (x - 1)(1 + x + x² + x³) = x⁴ - 1

So integrand becomes: 1 + x + x² + x³

Integrate: x + x²/2 + x³/3 + x⁴/4 + C

Answer: x + x²/2 + x³/3 + x⁴/4 + C

---

20. ∫ (x⁵ - 1)/(x - 1) dx

Similarly, (x⁵ - 1)/(x - 1) = 1 + x + x² + x³ + x⁴

Integrate: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + C

Answer: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + C

---

21. ∫ (x⁶ - 1)/(x - 1) dx

Same idea: = 1 + x + x² + x³ + x⁴ + x⁵

Integrate: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + C

Answer: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + C

---

22. ∫ (x⁷ - 1)/(x - 1) dx

= 1 + x + x² + ... + x⁶

Integrate: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + C

Answer: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + C

---

23. ∫ (x⁸ - 1)/(x - 1) dx

= 1 + x + ... + x⁷

Integrate: x + x²/2 + ... + x⁸/8 + C

Answer: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + x⁸/8 + C

---

24. ∫ (x⁹ - 1)/(x - 1) dx

= 1 + x + ... + x⁸

Integrate: x + x²/2 + ... + x⁹/9 + C

Answer: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + x⁸/8 + x⁹/9 + C

---

25. ∫ (x¹⁰ - 1)/(x - 1) dx

= 1 + x + ... + x⁹

Integrate: x + x²/2 + ... + x¹⁰/10 + C

Answer: x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + x⁸/8 + x⁹/9 + x¹⁰/10 + C

---

All done! Each problem was solved by simplifying first (factoring, expanding, or dividing), then integrating term by term.

Final Answer:
1. x⁵/5 + (4/3)x³ + 4x + C
2. (2/7)x^(7/2) - (2/3)x^(3/2) + 10√x + C
3. x⁶/6 + (3/2)x⁴ + 6x² + 8 ln|x| + C
4. x²/2 - x + C
5. -cos x + sin x + C
6. eˣ - e⁻ˣ + C
7. arctan(x) + C
8. tan x - cot x + C
9. x + ln|(x-1)/(x+1)| + C
10. x²/2 - (3/2) ln(x² + 1) + arctan(x) + C
11. x³/3 - x + C
12. x²/2 + x + C
13. x³/3 - x² + 4x + C
14. x²/2 + 2x + C
15. x³/3 + 4x + C
16. x³/3 + (3/2)x² + 9x + C
17. x²/2 + 2x + C
18. x³/3 - x²/2 + x + C
19. x + x²/2 + x³/3 + x⁴/4 + C
20. x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + C
21. x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + C
22. x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + C
23. x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + x⁸/8 + C
24. x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + x⁸/8 + x⁹/9 + C
25. x + x²/2 + x³/3 + x⁴/4 + x⁵/5 + x⁶/6 + x⁷/7 + x⁸/8 + x⁹/9 + x¹⁰/10 + C
Parent Tip: Review the logic above to help your child master the concept of u substitution worksheet.
Print Download

How to use

Click Print to open a print-ready version directly in your browser, or use Download to save the file to your device. The ⭐ Answer button generates an AI answer key instantly - useful for teachers who need a quick reference. Need a different version? Our AI Worksheet Generator lets you create a custom worksheet on any topic in seconds.

(view all u substitution worksheet)

Solved Substitution method Week 13 MAC 2311 Worksheet 1 1. | Chegg.com
Solved Worksheet on Integration Techniques 1. Evaluate (a) | Chegg.com
Integration Substitution Worksheet Solns | PDF | Mathematical ...
New Doc 2017-08-31 - Cal 2 Math 181 Worksheet Chapter 3 ...
integration by substitutions,when two function given,one is ...
Solved tice Worksheet on U-Substitution the integral. 1) | Chegg.com
integration by u substitution interactive worksheet
Quiz & Worksheet - U Substitution | Study.com
U-Substitution for Integration | Formula, Steps & Examples ...
Solved Worksheet on Integration Techniques 1. Evaluate (a) z ...