Integral Calculus Worksheets :: 25 - 48 - Free Printable
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Step-by-step solution for: Integral Calculus Worksheets :: 25 - 48
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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.
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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
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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
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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
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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
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5. ∫ (sin x + cos x) dx
Simple: ∫ sin x dx = -cos x, ∫ cos x dx = sin x
Answer: -cos x + sin x + C
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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
---
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.