Basic Calculus Worksheet No. 9 Indefinite Integration: r+2) R DR ... - Free Printable

Let’s solve each integral one by one. We’ll use basic integration rules, substitution where needed, and simplify step by step.

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1. ∫ 4x³√x dx

First, write √x as x^(1/2):

→ 4x³ * x^(1/2) = 4x^(7/2)

Now integrate:

∫ 4x^(7/2) dx = 4 * [x^(9/2) / (9/2)] + C = 4 * (2/9) x^(9/2) + C = (8/9)x^(9/2) + C

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2. ∫ y⁵(3y³ + 4) dy

Distribute y⁵:

= ∫ (3y⁸ + 4y⁵) dy

Integrate term by term:

= 3*(y⁹/9) + 4*(y⁶/6) + C = (1/3)y⁹ + (2/3)y⁶ + C

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3. ∫ (ax² + bx + c) dx

Integrate each term:

= a*(x³/3) + b*(x²/2) + c*x + C = (a/3)x³ + (b/2)x² + cx + C

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4. ∫ (∛x - 4/√x) dx

Write with exponents:

∛x = x^(1/3), 4/√x = 4x^(-1/2)

So: ∫ [x^(1/3) - 4x^(-1/2)] dx

Integrate:

= [x^(4/3)/(4/3)] - 4*[x^(1/2)/(1/2)] + C
= (3/4)x^(4/3) - 8x^(1/2) + C
= (3/4)x^(4/3) - 8√x + C

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5. ∫ ∜y (y³ - 2y) dy

∜y = y^(1/4)

Multiply: y^(1/4)*y³ = y^(13/4), y^(1/4)*(-2y) = -2y^(5/4)

So: ∫ [y^(13/4) - 2y^(5/4)] dy

Integrate:

= [y^(17/4)/(17/4)] - 2*[y^(9/4)/(9/4)] + C
= (4/17)y^(17/4) - (8/9)y^(9/4) + C
= (4/17)y^(17/4) - (8/9)y^(9/4) + C

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6. ∫ (x² + 4x - 3)/√x dx

Split into terms:

= ∫ [x²/x^(1/2) + 4x/x^(1/2) - 3/x^(1/2)] dx
= ∫ [x^(3/2) + 4x^(1/2) - 3x^(-1/2)] dx

Integrate:

= [x^(5/2)/(5/2)] + 4*[x^(3/2)/(3/2)] - 3*[x^(1/2)/(1/2)] + C
= (2/5)x^(5/2) + (8/3)x^(3/2) - 6x^(1/2) + C
= (2/5)x^(5/2) + (8/3)x^(3/2) - 6√x + C

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7. ∫ (-1/x² - 4/x + 8) dx

Rewrite: -x^(-2) - 4x^(-1) + 8

Integrate:

= -[x^(-1)/(-1)] - 4 ln|x| + 8x + C
= 1/x - 4 ln|x| + 8x + C
= 1/x - 4 ln|x| + 8x + C

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8. ∫ (x/4 - 4/x) dx

= (1/4)∫x⁴ dx - 4∫(1/x) dx
= (1/4)(x⁵/5) - 4 ln|x| + C
= (1/20)x⁵ - 4 ln|x| + C

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9. ∫ (4t² + 5)/∛(t²) dt

∛(t²) = t^(2/3)

So: (4t² + 5)/t^(2/3) = 4t^(4/3) + 5t^(-2/3)

Integrate:

= 4*[t^(7/3)/(7/3)] + 5*[t^(1/3)/(1/3)] + C
= (12/7)t^(7/3) + 15t^(1/3) + C
= (12/7)t^(7/3) + 15t^(1/3) + C

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10. ∫ ∜(4y - 9) dy

Let u = 4y - 9 → du = 4 dy → dy = du/4

Integral becomes: ∫ u^(1/4) * (du/4) = (1/4) ∫ u^(1/4) du
= (1/4) * [u^(5/4)/(5/4)] + C = (1/4)*(4/5)u^(5/4) + C = (1/5)u^(5/4) + C
Substitute back: (1/5)(4y - 9)^(5/4) + C

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11. ∫ 5/√(6y - 1) dy

Let u = 6y - 1 → du = 6 dy → dy = du/6

Integral: 5 ∫ u^(-1/2) * (du/6) = (5/6) ∫ u^(-1/2) du
= (5/6) * [u^(1/2)/(1/2)] + C = (5/6)*2*u^(1/2) + C = (5/3)√u + C
= (5/3)√(6y - 1) + C

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12. ∫ s ds / √(7s² + 4)

Let u = 7s² + 4 → du = 14s ds → s ds = du/14

Integral: ∫ (1/√u) * (du/14) = (1/14) ∫ u^(-1/2) du
= (1/14) * 2u^(1/2) + C = (1/7)√u + C
= (1/7)√(7s² + 4) + C

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13. ∫ (3 + s) (s + 1)² ds

Let u = 3 + s → s = u - 3 → s + 1 = u - 2

Then: ∫ √u * (u - 2)² du

Expand (u - 2)² = u² - 4u + 4

So: ∫ u^(1/2)(u² - 4u + 4) du = ∫ (u^(5/2) - 4u^(3/2) + 4u^(1/2)) du

Integrate:

= [u^(7/2)/(7/2)] - 4[u^(5/2)/(5/2)] + 4[u^(3/2)/(3/2)] + C
= (2/7)u^(7/2) - (8/5)u^(5/2) + (8/3)u^(3/2) + C

Substitute u = 3 + s:

= (2/7)(3+s)^(7/2) - (8/5)(3+s)^(5/2) + (8/3)(3+s)^(3/2) + C

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14. ∫ (x² + 2)² x³ dx

Let u = x² + 2 → du = 2x dx → but we have x³ dx = x² * x dx

Note: x² = u - 2, and x dx = du/2

So: x³ dx = x² * x dx = (u - 2) * (du/2)

Integral becomes: ∫ u² * (u - 2) * (du/2) = (1/2) ∫ (u³ - 2u²) du
= (1/2)[u⁴/4 - 2u³/3] + C = (1/8)u⁴ - (1/3)u³ + C

Substitute u = x² + 2:

= (1/8)(x²+2)^4 - (1/3)(x²+2)^3 + C

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15. ∫ t³ dt / √(1 - 2t²)

Let u = 1 - 2t² → du = -4t dt → t dt = -du/4

But we have t³ dt = t² * t dt

From u = 1 - 2t² → t² = (1 - u)/2

So: t³ dt = t² * t dt = [(1 - u)/2] * [-du/4] = -(1 - u)/8 du

Integral: ∫ [-(1 - u)/8] / √u du = (-1/8) ∫ (1 - u) u^(-1/2) du
= (-1/8) ∫ (u^(-1/2) - u^(1/2)) du
= (-1/8)[2u^(1/2) - (2/3)u^(3/2)] + C
= (-1/8)*2[u^(1/2) - (1/3)u^(3/2)] + C
= (-1/4)[√u - (1/3)u√u] + C
= (-1/4)√u [1 - u/3] + C

Substitute u = 1 - 2t²:

= (-1/4)√(1 - 2t²) [1 - (1 - 2t²)/3] + C

Simplify inside:

1 - (1 - 2t²)/3 = (3 - 1 + 2t²)/3 = (2 + 2t²)/3 = 2(1 + t²)/3

So:

= (-1/4)√(1 - 2t²) * [2(1 + t²)/3] + C
= (-1/6)(1 + t²)√(1 - 2t²) + C

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16. ∫ r² (r³/12 - 4)^5 dr

Let u = r³/12 - 4 → du = (3r²/12) dr = (r²/4) dr → r² dr = 4 du

Integral: ∫ u^5 * 4 du = 4 * (u^6 / 6) + C = (2/3)u^6 + C

Substitute back: (2/3)(r³/12 - 4)^6 + C

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17. ∫ 7x² / (2 - 5x³)^4 dx

Let u = 2 - 5x³ → du = -15x² dx → x² dx = -du/15

Integral: 7 ∫ u^(-4) * (-du/15) = (-7/15) ∫ u^(-4) du
= (-7/15) * [u^(-3)/(-3)] + C = (7/45) u^(-3) + C
= 7/(45(2 - 5x³)^3) + C

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18. ∫ (x² + 2x) / ∜(x³ + 3x² + 1) dx

Notice numerator is derivative of inside? Let’s check:

Let u = x³ + 3x² + 1 → du = (3x² + 6x) dx = 3(x² + 2x) dx → so (x² + 2x) dx = du/3

Denominator: ∜u = u^(1/4)

Integral: ∫ u^(-1/4) * (du/3) = (1/3) ∫ u^(-1/4) du
= (1/3) * [u^(3/4)/(3/4)] + C = (1/3)*(4/3)u^(3/4) + C = (4/9)u^(3/4) + C
= (4/9)(x³ + 3x² + 1)^(3/4) + C

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19. ∫ (8x - 2) √(3 - x + 2x²) dx

Let u = 3 - x + 2x² → du = (-1 + 4x) dx = (4x - 1) dx

But we have (8x - 2) = 2*(4x - 1) → perfect!

So: (8x - 2) dx = 2 du

Integral: ∫ u * 2 du = 2 ∫ u^(1/2) du = 2*(2/3)u^(3/2) + C = (4/3)u^(3/2) + C
= (4/3)(3 - x + 2x²)^(3/2) + C

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20. ∫ sin(4x) / (2 + cos(4x))² dx

Let u = 2 + cos(4x) → du = -4 sin(4x) dx → sin(4x) dx = -du/4

Integral: ∫ (1/u²) * (-du/4) = (-1/4) ∫ u^(-2) du
= (-1/4) * [u^(-1)/(-1)] + C = (1/4) u^(-1) + C
= 1/[4(2 + cos(4x))] + C

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21. ∫ csc²(3x) (1 - cot(3x))^9 dx

Let u = 1 - cot(3x) → du = -[-csc²(3x) * 3] dx = 3 csc²(3x) dx → csc²(3x) dx = du/3

Integral: ∫ u^9 * (du/3) = (1/3) * (u^10 / 10) + C = (1/30)u^10 + C
= (1/30)(1 - cot(3x))^10 + C

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22. ∫ (∛r + 2)^(2/3) / ∛(r²) dr

Write: ∛r = r^(1/3), ∛(r²) = r^(2/3)

So integrand: (r^(1/3) + 2)^(2/3) / r^(2/3) = (r^(1/3) + 2)^(2/3) * r^(-2/3)

Let u = r^(1/3) + 2 → then r^(1/3) = u - 2 → r = (u - 2)^3

dr = 3(u - 2)^2 du

Also, r^(-2/3) = [(u - 2)^3]^(-2/3) = (u - 2)^(-2)

So integral becomes:

∫ u^(2/3) * (u - 2)^(-2) * 3(u - 2)^2 du = 3 ∫ u^(2/3) du
= 3 * [u^(5/3)/(5/3)] + C = 3*(3/5)u^(5/3) + C = (9/5)u^(5/3) + C

Substitute u = r^(1/3) + 2:

= (9/5)(r^(1/3) + 2)^(5/3) + C

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23. ∫ x(x² + 1) ∜(4 - 2x² - x⁴) dx

Note: 4 - 2x² - x⁴ = 4 - (x⁴ + 2x²) = 4 - (x² + 1)^2 + 1? Wait, let's try substitution.

Let u = x² + 1 → du = 2x dx → x dx = du/2

Also, x⁴ + 2x² = (x² + 1)^2 - 1 = u² - 1

So 4 - 2x² - x⁴ = 4 - (x⁴ + 2x²) = 4 - (u² - 1) = 5 - u²

Wait, original expression: 4 - 2x² - x⁴ = -(x⁴ + 2x² - 4) — not matching.

Actually: 4 - 2x² - x⁴ = 4 - (x⁴ + 2x²) = 4 - [(x² + 1)^2 - 1] = 4 - (u² - 1) = 5 - u²

Yes!

And we have x(x² + 1) dx = x * u * dx = u * (x dx) = u * (du/2)

So integral: ∫ u * ∜(5 - u²) * (du/2) = (1/2) ∫ u (5 - u²)^(1/4) du

Now let v = 5 - u² → dv = -2u du → u du = -dv/2

Integral: (1/2) ∫ v^(1/4) * (-dv/2) = (-1/4) ∫ v^(1/4) dv
= (-1/4) * [v^(5/4)/(5/4)] + C = (-1/4)*(4/5)v^(5/4) + C = (-1/5)v^(5/4) + C

Substitute v = 5 - u² = 5 - (x² + 1)^2

= (-1/5)[5 - (x² + 1)^2]^(5/4) + C

But 5 - (x² + 1)^2 = 5 - (x⁴ + 2x² + 1) = 4 - 2x² - x⁴ — which matches original!

So final answer: (-1/5)(4 - 2x² - x⁴)^(5/4) + C

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24. ∫ x dx / √(x + 4)

Let u = x + 4 → x = u - 4, dx = du

Integral: ∫ (u - 4)/√u du = ∫ (u^(1/2) - 4u^(-1/2)) du
= [u^(3/2)/(3/2)] - 4[u^(1/2)/(1/2)] + C
= (2/3)u^(3/2) - 8u^(1/2) + C
= (2/3)(x+4)^(3/2) - 8√(x+4) + C
= (2/3)(x+4)^(3/2) - 8√(x+4) + C

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25. ∫ √(1/t - 1) * dt / t²

Let u = 1/t - 1 → du = -1/t² dt → dt/t² = -du

Integral: ∫ √u * (-du) = - ∫ u^(1/2) du = - (2/3) u^(3/2) + C
= -(2/3)(1/t - 1)^(3/2) + C

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26. ∫ √(3 - 2x) * x² dx

Let u = 3 - 2x → x = (3 - u)/2, dx = -du/2

x² = [(3 - u)/2]^2 = (9 - 6u + u²)/4

Integral: ∫ √u * (9 - 6u + u²)/4 * (-du/2) = (-1/8) ∫ u^(1/2)(9 - 6u + u²) du
= (-1/8) ∫ (9u^(1/2) - 6u^(3/2) + u^(5/2)) du
= (-1/8)[9*(2/3)u^(3/2) - 6*(2/5)u^(5/2) + (2/7)u^(7/2)] + C
= (-1/8)[6u^(3/2) - (12/5)u^(5/2) + (2/7)u^(7/2)] + C
= (-6/8)u^(3/2) + (12/40)u^(5/2) - (2/56)u^(7/2) + C
= (-3/4)u^(3/2) + (3/10)u^(5/2) - (1/28)u^(7/2) + C

Substitute u = 3 - 2x:

= (-3/4)(3-2x)^(3/2) + (3/10)(3-2x)^(5/2) - (1/28)(3-2x)^(7/2) + C

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27. ∫ (1 + √x)^6 / √x dx

Let u = 1 + √x → du = (1/(2√x)) dx → dx/√x = 2 du

Integral: ∫ u^6 * 2 du = 2*(u^7 / 7) + C = (2/7)u^7 + C
= (2/7)(1 + √x)^7 + C

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28. ∫ r(2r² + 1)^5 dr

Let u = 2r² + 1 → du = 4r dr → r dr = du/4

Integral: ∫ u^5 * (du/4) = (1/4)*(u^6 / 6) + C = (1/24)u^6 + C
= (1/24)(2r² + 1)^6 + C

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29. ∫ cos⁴(3y) sin(3y) dy

Let u = cos(3y) → du = -3 sin(3y) dy → sin(3y) dy = -du/3

Integral: ∫ u^4 * (-du/3) = (-1/3) ∫ u^4 du = (-1/3)(u^5 / 5) + C = (-1/15)u^5 + C
= (-1/15)cos⁵(3y) + C

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30. ∫ sec²y tan y dy

Let u = tan y → du = sec²y dy

Integral: ∫ u du = u²/2 + C = (1/2)tan²y + C

Alternatively, could let u = sec y → du = sec y tan y dy, then integral becomes ∫ u du = same thing? Wait:

If u = sec y, then du = sec y tan y dy, but we have sec²y tan y dy = sec y * (sec y tan y dy) = u du → same result: u²/2 + C = (1/2)sec²y + C

But note: (1/2)tan²y and (1/2)sec²y differ by constant since sec²y = tan²y + 1. Both are correct, but usually we pick simpler form. Since the problem has tan y, maybe keep tan. But both acceptable. However, standard answer often uses tan.

Actually, let’s verify derivative:

d/dy [ (1/2)tan²y ] = (1/2)*2 tan y * sec²y = tan y sec²y ✓

d/dy [ (1/2)sec²y ] = (1/2)*2 sec y * sec y tan y = sec²y tan y ✓

Same thing! So either is fine. But since the integrand has tan y, perhaps (1/2)tan²y is more direct.

We'll go with: (1/2)tan²y + C

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Final Answer:

1. (8/9)x^(9/2) + C
2. (1/3)y⁹ + (2/3)y⁶ + C
3. (a/3)x³ + (b/2)x² + cx + C
4. (3/4)x^(4/3) - 8√x + C
5. (4/17)y^(17/4) - (8/9)y^(9/4) + C
6. (2/5)x^(5/2) + (8/3)x^(3/2) - 6√x + C
7. 1/x - 4 ln|x| + 8x + C
8. (1/20)x⁵ - 4 ln|x| + C
9. (12/7)t^(7/3) + 15t^(1/3) + C
10. (1/5)(4y - 9)^(5/4) + C
11. (5/3)√(6y - 1) + C
12. (1/7)√(7s² + 4) + C
13. (2/7)(3+s)^(7/2) - (8/5)(3+s)^(5/2) + (8/3)(3+s)^(3/2) + C
14. (1/8)(x²+2)^4 - (1/3)(x²+2)^3 + C
15. (-1/6)(1 + t²)√(1 - 2t²) + C
16. (2/3)(r³/12 - 4)^6 + C
17. 7/[45(2 - 5x³)^3] + C
18. (4/9)(x³ + 3x² + 1)^(3/4) + C
19. (4/3)(3 - x + 2x²)^(3/2) + C
20. 1/[4(2 + cos(4x))] + C
21. (1/30)(1 - cot(3x))^10 + C
22. (9/5)(r^(1/3) + 2)^(5/3) + C
23. (-1/5)(4 - 2x² - x⁴)^(5/4) + C
24. (2/3)(x+4)^(3/2) - 8√(x+4) + C
25. -(2/3)(1/t - 1)^(3/2) + C
26. (-3/4)(3-2x)^(3/2) + (3/10)(3-2x)^(5/2) - (1/28)(3-2x)^(7/2) + C
27. (2/7)(1 + √x)^7 + C
28. (1/24)(2r² + 1)^6 + C
29. (-1/15)cos⁵(3y) + C
30. (1/2)tan²y + C