Here we provide RD Sharma Class 12 Ex 19.2 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.2 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.
Textbook | NCERT |
Class | Class 12th |
Subject | Maths |
Chapter | 19 |
Exercise | 19.2 |
Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 19.2 Solutions Chapter 19 Indefinite Integrals
Question 1. Evaluate (∫(3x√5 + 4√x + 5)dx
Solution:
We have, (∫(3x√5+4√x+5)dx
= ∫3x√5 dx + ∫4√x dx + ∫5dx
= ∫3x3/2dx + 4∫x1/2 dx + 5∫dx
= x(3/2)+1/(3/2 + 1) + 4x(1/2)+1/(1/2 + 1) + 5x + c
= 6/5 x5/2 + 8/3 x3/2 + 5x + c
Question 2. Evaluate ∫(2x + 5/x – 1/x1/3)dx
Solution:
We have, ∫(2x + 5/x – 1/x1/3)dx
= ∫2xdx + 5∫1/x dx – ∫1/x1/3 dx
= 2x/(log2) + 5logx – 3/2 x2/3 + c
Question 3. Evaluate ∫{√x (ax2 + bx + c)}dx
Solution:
We have, ∫{√x (ax2 + bx + c)}dx
= ∫√x × ax2 dx + ∫√x × bx dx + ∫c√x dx
= ∫ax5/2 dx + ∫bx3/2dx + ∫cx1/2 dx
= (ax(5/2)+1)/(5/2 + 1) + (bx(3/2)+1)/(3/2 + 1) + (cx(1/2)+1)/(1/2 + 1) + d
= (2ax7/2)/7 + (2bx5/2)/5 + (2cx3/2)/3 + d
Question 4. Evaluate ∫(2 – 3x) (3 + 2x)(1 – 2x)dx
Solution:
We have, ∫(2 – 3x) (3 + 2x)(1 – 2x)dx
= ∫(6 + 4x – 9x – 6x2)(1 – 2x)dx
= ∫(-6x2 – 5x + 6)(1 – 2x)dx
= ∫(-6x2 + 12x3 – 5x + 10x2 + 6 – 12x)dx
= ∫(4x2 + 12x3 – 17x + 6)dx
= ∫(12x3 + 4x2 – 17x + 6)dx
= 12/4 x4 + 4/3 x3 – 17/2 x2 + 6x + c
= 3x4 + 4/3 x3 – 17/2 x2 + 6x + c
Question 5. Evaluate ∫(m/x + x/m + mx + xm + mx)dx
Solution:
We have, ∫(m/x + x/m + mx + xm + mx)dx
= m∫1/x dx + 1/m ∫xdx + ∫mxdx + ⌋xmdx + m∫xdx
= mlog|x| + x2/2m + mx/(logm) + xm+1/(m + 1) + (mx2)/2 + c
Question 6. Evaluate ∫ (√x – 1/√x)2 dx
Solution:
We have, ∫ (√x – 1/√x)2 dx
By using formula (x + y)2 = x2 + y2 +2xy
We get, ∫(x + 1/x – 2)dx
= ∫xdx + ∫1/x dx – 2∫1.dx
= x2/2 + log|x| – 2x + C
Question 7. Evaluate ∫((1 + x)3)/√xdx
Solution:
We have, ∫((1 + x)3)/√xdx
By using formula (x + y)3 = x3 + y3 +3x2y + 3xy2
We get, ∫(1 + x3 + 3x2 + 3x)/√x dx
= 1/√x dx + ∫x3/√x dx + ∫(3x2)/√x dx + ∫3x/√x dx
= ∫x-1/2 dx + ∫x5/2 dx + 3∫x3/2 dx + 3∫x1/2 dx
= x(-1/2)+1/((-1)/2 + 1) + (x(5/2)+1)/(5/2 + 1) + (3x(3/2)+1)/(3/2 + 1) + 3 x(1/2)+1/(1/2 + 1) + c
= x1/2/(1/2) + x7/2/(7/2) + (3x5/2)/(5/2) + 3 x3/2/(3/2) + c
= 2x1/2 + 2/7 x7/2 + 6/5 x5/2 + 6/3 x3/2 + c
= 2x1/2 + 2/7 x7/2 + 6/5 x5/2 + 2x3/2 + c
Question 8. Evaluate ∫{x2 + elogx + (e/2)x }dx
Solution:
We have, ∫{x2 + elogx + (e/2)x }dx
= ∫x2 dx + ∫elogx dx + ∫(e/2)x dx
= x3/3 + ∫xdx + ∫(e/2)xdx
= x3/3 + x2/2 + 1/(log(e/2)) × (e/2)x + c
Question 9. Evaluate ∫ (xe + ex + ee)dx
Solution:
We have, ∫ (xe + ex + ee)dx
= ∫xe dx + ∫exdx + ∫eedx
= xe+1/(e + 1) + ex + eex + c
Question 10. Evaluate ∫√x (x3 – 2/x)dx
Solution:
We have, ∫√x (x3 – 2/x)dx
= ∫ x7/2 dx – 2∫ x-1/2 dx
= x(7/2)+1/(7/2 + 1) – 2 x(-1/2)+1/((-1)/2 + 1) + c
= x9/2/(9/2) – (2x-1/2)/((-1)/2) + c
= 2/9 x9/2 – 4x-1/2 + c
Question 11. Evaluate ∫1/√x (1 + 1/x)dx
Solution:
We have, ∫1/√x (1 + 1/x)dx
= ∫ (1/√x + 1/(√x × x))dx
= ∫x-1/2 + ∫x-3/2 dx
= 2x1/2 – 2x-1/2 + c
Question 12. Evaluate ∫(x6 + 1)/(x2 + 1) dx
Solution:
We have, ∫(x6 + 1)/(x2 + 1) dx
= ∫((x2)3 + (1)3)/(x2 + 1) dx
= ∫(x2 + 1)(x4 + 1 – x2)/(x2 + 1) dx
= ∫(x4 – x2 + 1)dx
= ∫x4dx – ∫x2dx + ∫1dx
= x5/5 – x3/3 + x + c
Question 13. Evaluate ∫ (x-1/3 + √x + 2)/∛x dx
Solution:
We have, (x-1/3 + √x + 2)/∛x dx
= ∫(x-1/3 dx)/x1/3 + ∫x1/2/x1/3dx + ∫2/x1/3dx
= ∫x-2/3 dx + ∫x1/6 dx + 2∫ x-1/3dx
= 3x1/3 + 6/7 x7/6 + 3x2/3 + c
Question 14. Evaluate ∫((1 + √x)2)/√x dx
Solution:
We have, ∫((1 + √x)2)/√x dx
= ∫(1 + x + 2√x)/x1/2dx
= ∫x-1/2+∫ x1/2 dx + 2∫dx
= 2x1/2 + 2/3 x3/2 + 2x + c
= 2√x + 2x + 2/3 x3/2 + c
Question 15. Evaluate ∫√x(3 – 5x)dx
Solution:
We have, ∫√x(3 – 5x)dx
= 3∫√x dx – 5x3/2 dx
= 3x3/2/(3/2) – 5 x5/2/(5/2) + c
= 2x3/2 – 2x5/2 + c
Question 16. Evaluate ∫((x + 1)(x – 2))/√x dx
Solution:
We have, ∫((x + 1)(x – 2))/√x dx
= ∫(x2 – 2x + x – 2)/x1/2dx
= ∫(x2 – x – 2)/x1/2dx
= ∫x2/x1/2dx – ∫x1/2dx – 2∫x-1/2dx
= (2x5/2)/5 – (2x3/2)/3 – 4x1/2 + c
= 2/5 x5/2 – (2x3/2)/3 – 4√x + c
Question 17. Evaluate ∫(x5 + x-2 + 2)/x2dx
Solution:
We have, ∫(x5 + x-2 + 2)/x2dx
= ∫(x5/x2 +x-2/x2 +2/x2)dx
= ∫x3dx + ∫x-4 + 2∫x-2 dx
= x4/4 + x-3/(-3) + (2x-1)/(-1) + c
= x4/4 – x-3/3 – 2/x + c
Question 18. Evaluate ∫(3x + 4)2 dx
Solution:
We have, ∫(3x + 4)2 dx
By using formula (x + y)2 = x2 + y2 +2xy
We get, ∫ (9x2 + 16 + 24x)dx
= ∫9x2 dx + ∫16dx + ∫24xdx
= 9 x3/3 + 16x + 24 x2/2 + c
= 3x3 + 16x + 12x2 + c
Question 19. Evaluate ∫(2x4 + 7x3 + 6x2)/(x2 + 2x) dx
Solution:
We have, ∫(2x4 + 7x3 + 6x2)/(x2 + 2x) dx
= ∫x(2x3 + 7x2 + 6x)/(x(x + 2))dx
= ∫(2x3 + 7x2 + 6x)/(x + 2)dx
= ∫ (2x3 + 4x2 + 3x2 + 6x)/((x + 2))dx
= ∫(2x2(x + 2) + 3x(x + 2))/(x + 2) dx
= ∫(x + 2)(2x2 + 3x)/(x + 2) dx
= ∫(2x2 + 3x)dx
= ∫2x2 dx + ∫3xdx
= 2/3 x3 + 3/2 x2 + c
Question 20. Evaluate ∫(5x4 + 12x3 + 7x2)/(x2 + x) dx
Solution:
We have, ∫(5x4 + 7x3 + 5x3 + 7x2)/(x2 + x) dx
= ∫(5x3 + 7x2 + 5x2 + 7x)/(x + 1) dx
= ∫5x2 (x + 1) + 7x(x + 1)/(x + 1) dx
= ∫(5x2 + 7x)dx
= (5x3)/3 + (7x2)/2 + C
Question 21. Evaluate ∫(sin2x)/(1 + cosx) dx
Solution:
We have, ∫(sin2x)/(1 + cosx) dx
= ∫(1 – cos2x)/(1 + cosx) dx
= ∫((1 – cosx)(1 + cosx))/(1 + cosx) dx
= ∫(1 – cosx)dx
= x – sinx + c
Question 22. Evaluate ∫(sec2x + cosec2 x)dx
Solution:
We have, ∫(sec2x + cosec2 x)dx
= ∫ sec2xdx + ∫ cosec2xdx
= tanx – cotx + c
= tanx – cotx + c
Question 23. Evaluate ∫(sin3x – cos3x)/(sin2xcos2x) dx
Solution:
We have, ∫(sin3x – cos3x)/(sin2xcos2x) dx
= ∫((sin3x)/(sin2x cos2x) – (cos3x)/(sin2x cos2x))dx
= ∫ (sinxsec2x – cosxcosec2x)dx
= ∫ (tanxsecx – cotxcosecx)dx
= secx + cosecx + c
Question 24. Evaluate ∫(5cos3x + 6sin3x)/(2sin2xcos2x) dx
Solution:
We have, ∫(5cos3x + 6sin3x)/(2sin2xcos2x) dx
= ∫(5cos3x)/(2sin2xcos2x) dx + ∫(6sin3x)/(2sin2xcos2x) dx
= 5/2 ∫(cosx)/(sin2x) dx + 3∫(sinx)/(cos2x) dx
= 5/2 ∫cotxcosecxdx + 3∫ secxtanxdx
= (-5)/2 cosecx + 3secx+c
= (-5)/2 cossecx + 3secx+c
Question 25. Evaluate ∫(tanx + cotx)2 dx
Solution:
We have, ∫(tanx + cotx)2 dx
By using formula (x + y)2 = x2 + y2 + 2xy
We get, ∫(tan2x + cot2x + 2tanx cotx)dx
= ∫ (sec2x – 1 + cosec2x – 1 + ((2 × 1)/cotx) × cotx)dx
= ∫ (sec2x + cosec2x)dx
= ∫sec2xdx + ∫cosec2xdx
= tanx – cotx + c
Question 26. Evaluate ∫(1 – cos2x)/(1 + cos2x) dx
Solution:
We have, ∫(1 – cos2x)/(1 + cos2x) dx
= ∫(2sin2x)/(2cos2x) dx
= ∫tan2xdx
= ∫(sec2x – 1)dx
= ∫sec2xdx – 1∫dx
= tanx – x + c
Question 27. Evaluate ∫(cosx)/(1 – cosx) dx
Solution:
We have, ∫(cosx)/(1 – cosx) dx
= ∫(cosx(1 + cosx))/((1 – cosx)(1 + cosx)) dx
= ∫(cosx + cos2x)/(1 – cos2x) dx
= ∫(cosx + cos2x)/(sin2x) dx
= ∫(cosx)/(sin2x) dx + ∫(cos2x)/(sin2x) dx [Since, cosx/sinx = cotx]
= ∫cotx × cosecxdx + ∫(cosec2x – 1)dx [Since, cot2x = cosec2x – 1]
= -cosecx – cotx – x + c
Question 28. Evaluate ∫cos2x – sin2x/√(1 + cos4x) dx
Solution:
We have, ∫cos2x – sin2x/√(1 + cos4x) dx
= ∫(cos2x – sin2x)/√(2cos22x) dx
= 1/√2 ∫(cos2x – sin2x)/(cos2x) dx
= 1/√2∣(cos2x – sin2x)/(cos2x – sin2x) dx
= 1/√2∫1 × dx
= x/√2 + c
Question 29. Evaluate ∫ 1/(1 – cosx) dx
Solution:
We have, ∫ 1/(1 – cosx) dx
= ∫1/(1 – cosx) × (1 + cosx)/(1 + cosx) × dx
= ∫(1 + cosx)/(1 – cos2x) × dx
= ∫(1 + cosx)/(sin2x) × dx
= ∫1/(sin2x) dx + ∫(cosx)/(sin22x) dx
= ∫cosec2xdx + ∫cotx × cosecx dx
= -cotx – cosecx + c
Question 30. Evaluate ∫1/(1 – sinx) dx
Solution:
We have, ∫1/(1 – sinx) dx
= ∫1/(1 – sinx) × (1 + sinx)/(1 + sinx) × dx
= ∫(1 + sinx)/(1 – sin2x) × dx
= ∫(1 + sinx)/(cos2x) × dx
= ∫(1/(cos2x) + (sinx)/(cos2x)) × dx
= ∫1/(cos2x) dx + ∫(sinx)/(cos2x) × dx
= ∫sec2xdx + ∫tanx secx dx
= tanx + secx + c
Question 31. Evaluate ∫(tanx)/(secx + tanx) dx
Solution:
We have, ∫(tanx)/(secx + tanx) dx
= ∫(tanx)/(secx + tanx) × (secx – tanx)/(secx – tanx) × dx
= ∫(tanx(secx – tanx))/(sec2x – tan2x) × dx
= ∫(tanxsecx – tan2x)dx
= ∫sectanxdx – ∫(sec2x – 1)dx
= ∫secxtanxdx – ∫sec2xdx + 1∫dx
= secx – tanx + x + c
Question 32. Evaluate ∫(cosecx)/(cosecx – cotx)dx
Solution:
We have, ∫(cosecx)/(cosecx – cotx)dx
= ∫(cosecx)/(cosecx – cotx) × (cosecx + cotx)/(cosecx + cotx) × dx
= ∫(cosecx(cosecx + cotx))/(cosec2x – cot2x) × dx
= ∫(cosec2x + cosecx cotx)dx
= ∫cosec2xdx + ∫cosecx cotx dx
= -cotx – cosecx + c
Question 33. Evaluate ∫1/(1 + cos2x) dx
Solution:
We have, ∫1/(1 + cos2x) dx
= ∫ 1/(2cos2x) × dx
= 1/2 ∫sec2x × dx
= 1/2 × tanx + c
= (tanx)/2 + c
Question 34. Evaluate∫1/(1 – cos2x) dx
Solution:
We have, ∫1/(1 – cos2x) dx
= ∫1/(2sin2x)dx
= 1/2 ∫cosec2x dx
= (-1)/2 × cotx + c
= (-cotx)/2 + c
Question 35. Evaluate ∫tan-1[(sin2x)/(1 + cos2x)]dx
Solution:
We have, ∫tan-1[(sin2x)/(1 + cos2x)]dx
= ∫tan-1[(2sinxcosx)/(2cos2x)]dx
= ∫tan-1[(sinx)/(cosx)]dx
= ∫tan-1(tanx)dx
= ∫xdx
= x2/2 + c
Question 36. Evaluate ∫cos-1(sinx)dx
Solution:
We have, ∫cos-1(sinx)dx
= ∫cos-1[cos(π/2 – x)]dx
= ∫(π/2 – x)dx
= π/2 ∫dx – ∫xdx
= π/2 × x – x2/2 + c
Question 37. Evaluate ∫ cot-1(sinx)dx
Solution:
We have, ∫ cot-1(sinx)dx
= ∫cot-1[(sin2x)/(1 – cos2x)]dx
= ∫cot-1((cosx)/(sinx))dx
= ∫cot-1(cotx)dx
= ∫xdx
= x2/2 + c
Question 38. Evaluate ∫ sin-1((2tanx)/(1 + tan2x))dx
Solution:
We have, ∫ sin-1((2tanx)/(1 + tan2x))dx
= ∫ sin-1(sin2x)dx
= ∫2xdx
= 2∫xdx
= (2x2)/2 + c
= x2 + c
Question 39. Evaluate ∫((x3 + 8)(x – 1))/(x2 – 2x + 4) dx
Solution:
We have, ∫((x3 + 8)(x – 1))/(x2 – 2x + 4) dx
= ∫((x + 2)(x2 – 2x + 4)(x – 1))/(x2 – 2x + 4) dx
= ∫(x + 2)(x – 1)dx
= ∫(x2 – x+2x – 2)dx
= ∫(x2 + x – 2)dx
= x3/3 + x2/2 – 2x + c
Question 40. Evaluate ∫(atanx + bcotx)2 dx
Solution:
We have, ∫(atanx + bcotx)2 dx
By using formula (x + y)2 = x2 + y2 + 2xy , we get
= ∫(a2 tan2x + b2cot2x + 2ab tanx cotx)dx
= ∫[a2 (sec2x – 1) + b2(cosec2x – 1) + 2ab]dx
= ∫[a2 sec2x – a2 + b2cosec2x – b2 + 2ab]dx
= a2tanx – a2x – b2 cotx – b2x + 2abx + c
= a2tanx – b2 cotx – (a2 + b2 – 2ab)x + c
Question 41. Evaluate ∫(x3 – 3x2 + 5x – 7 + x2 ax)/(2x2) dx
Solution:
We have, ∫(x3 – 3x2 + 5x – 7 + x2 ax)/(2x2) dx
= 1/2 ∫x3/x2dx – 3/2∫x2/x2dx + 5/2∫x/x2dx – 7/2∫x-2dx + 1/2∫(x2ax)/x2dx
= 1/2 × x2/2 – 3/2x + 5/2 logx – 7/2 x-1 + 1/2ax/(loga) + c
= 1/2 [x2/2 – 3x + 5logx + 7/x + ax/(loga)] + c
Question 42. Evaluate ∫cosx/(1 + cosx) dx
Solution:
We have, ∫cosx/(1 + cosx) dx …..(1)
Now solve![]()
Since, cosx = cos2x/2 – sin2x/2 and cosx + 1 = 2cos2x/2
So, we get cosx/(1 + cosx) = 1/2[1 – tan2x/2]
Now put this value in eq(1), we get
= 1/2 ∫(1 – tan2x/2)dx
= 1/2 ∫(1 – sec2x/2 + 1)dx
= 1/2 ∫(2 – sec2x/2)dx
= 1/2 [2x – (tanx/2)/(1/2)] + c
= x – tanx/2 + c
Question 43. Evaluate∫(1 – cosx)/(1 + cosx) dx
Solution:
We have, ∫(1 – cosx)/(1 + cosx) dx ….(1)
Now solve
(1 – cosx)/(1 + cosx) = (2sin2x)/(2cos2x)
= tan2x/2
= (sec2x/2 – 1) [Since, 2sin2x/2 = 1 – cosx and 2cos2x/2 = 1 + cosx]
Now put this value in eq(1), we get
= ∫(sec2x/2 – 1)dx
= tan(x/2)/(1/2) – x + c
= 2tanx/2 – x + c
Question 44. Evaluate ∫{3sinx – 4cosx + 5/(cos2x) – 6/(sin2x) + tan2x – cot2x}dx
Solution:
We have, ∫{3sinx – 4cosx + 5/(cos2x) – 6/(sin2x) + tan2x – cot2x}dx
= 3∫sinxdx – 4∫cosxdx + 5∫sec2dx – 6∫cosec2x + ∫tan2xdx – ∫cot2xdx
= 3∫sinxdx – 4∫cosxdx + 5∫sec2xdx – 6∫cosec2x + ∫(sec2x – 1)dx – ∫(cosec2x – 1)dx
= 3∫sinxdx – 4∫cosxdx + 6∫sec2xdx – 7∫cosec2xdx
= -3cosx – 4sinx + 6tanx + 7cotx + c
Question 45. If f'(x) = x – 1/x2 and f(1) = 1/2, find f(x)?
Solution:
Given that ∫f'(x) = x – 1/x2
and f(1) = 1/2
We have to find f(x)
So, ∫f'(x) = ∫xdx – ∫1/x2dx
f(x) = x2/2 + x-1 + c
f(x) = x2/2 + 1/x + c
f(x) = x2/2 + 1/x + c …..(i)
As we know that
f(1) = 1/2
12/2 + 1/1 + c = 1/2
1/2 + 1 + c = 1/2
c = -1
On putting c = -1 in (i), we get
f(x) = x2/2 + 1/x – 1
Question 46. If f'(x) = x + b, f(1) = 5, f(2) = 13, find f(x)?
Solution:
Given that f'(x) = x + b
and f(1) = 5, f(2) = 13
We have to find f(x)
So, ∫f'(x) = ∫(x + b)dx
f(x) = x2/2 + bx + c …….(i)
As we know that
f(1) = 5
12/2 + b × 1 + c = 5
1/2 + b + c = 5
b + c = 9/2 …….(ii)
Also, f(2) = 13
22/2 + b × 2 + c = 13
2 + 2b + c = 13
2b + c = 11 …….(iii)
Now, subtract eq(ii) from eq(iii), we get
b = 11 – 9/2
b = 13/2
Now, put b = 13/2 in eq(ii), we get
13/2 + c = 9/2
c = 9/2 – 13/2
c = (9 – 13)/2
= (-4)/2
= -2
Now, on putting b = 13/2 and c = -2 in equation (i), we get
f(x) = x2/x + 13/2x – 2
f(x) = x2/2 + 13/2x – 2
Question 47. If f'(x) = 8x3 – 2x, f(2) = 8, find f(x)?
Solution:
Given that f'(x) = 8x3 – 2x
and f(2) = 8
We have to find f(x)
So, ∫f'(x)dx = ∫(8x3 – 2x)dx
f(x) = ∫(8x3 – 2x)dx
= ∫8x3dx – ∫2xdx
= (8x4)/4 – (2x2)/2 + c
= 2x4 – x2 + c
f(x) = 2x4 – x2 + c ……….(i)
As we know that f(2) = 8
So, f(2) = 2(2)4 – (2)2 + c = 8
32 – 4 + c = 8
28 + c = 8
c = -20
Now, Put c = -20 in eq(i), we get
f(x) = 2x4 – x2 – 20
Question 48. If f'(x) = asinx + bcosx and f'(0) = 4, f(0) = 3, f(π/2) = 5, find f(x)?
Solution:
Given that, f'(x) = asinx + bcosx
and f'(0) = 4, f(0) = 3, f(π/2) = 5
We have to find f(x)
So,
∫f'(x) = ∫(asinx + bcosx)dx
f(x) = -acosx + bsinx + c
f(x) = -acosx + bsinx + c ………(i)
As we know that f'(0) = 4
So, f'(0) = asin0 + bcos0 = 4
a × 0 + b × 1 = 4
b = 4
Also, f(0) = 3
f(0) = -acos0 + bsin0 + c = 3
-a + 0 + c = 3
c – a = 3 ……..(ii)
Also, f(π/2) = 5
f(π/2) = -acos(π/2) + bsin(π/2) + c = 5
-a × 0 + b × 1 + c = 5
b + c = 5
4 + c = 5 [Since, b = 4]
c = 5 – 4
c = 1
Now, put c = 1 in eq(ii), we get 1 – a = 3
-a = 3 – 1
-a = 2
a = -2
Now, put a = -2, b = 4, and c = 1 in eq(i), we get
f(x) = -(-2)cosx + 4sinx + 1
f(x) = 2cosx + 4sinx + 1
Question 49. Write the primitive or anti-derivative of f(x) = √x + 1/√x.
Solution:
We have, f(x) = √x + 1/√x
∫f(x) = ∫(√x + 1/√x)dx
= ∫x1/2dx + ∫ x-1/2 dx
= 2/3 x3/2 + 2x1/2 + c
Hence, the primitive or anti-derivative of f(x) is 2/3 x3/2 + 2x1/2 + c.
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