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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.2 | Set 1

  • Last Updated : 28 Apr, 2021

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/(log⁡2) + 5log⁡x – 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/(log⁡m) + 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 + cos⁡x) dx

Solution:

We have, ∫(sin2x)/(1 + cos⁡x) dx

= ∫(1 – cos2⁡x)/(1 + cos⁡x) dx

= ∫((1 – cos⁡x)(1 + cos⁡x))/(1 + cos⁡x) dx

= ∫(1 – cos⁡x)dx

= x – sin⁡x + c

Question 22. Evaluate ∫(sec2x + cos⁡ec2 x)dx

Solution:

We have, ∫(sec2x + cos⁡ec2 x)dx

= ∫ sec2xdx + ∫ cosec2⁡xdx



= tan⁡x – cot⁡x + c

 = tan⁡x – cot⁡x + c

Question 23. Evaluate ∫(sin3⁡x – cos3x)/(sin2⁡xcos2x) dx

Solution:

We have, ∫(sin3⁡x – cos3x)/(sin2⁡xcos2x) dx

= ∫((sin3⁡x)/(sin⁡2x cos2⁡x) – (cos3x)/(sin⁡2x cos2x))dx 

= ∫ (sin⁡xsec2x – cos⁡xcos⁡ec2x)dx 

= ∫ (tan⁡xsec⁡x – cot⁡xcos⁡ecx)dx

= sec⁡x + cosec⁡x + c

Question 24. Evaluate ∫(5cos3⁡x + 6sin3x)/(2sin2xcos2x) dx

Solution:

We have, ∫(5cos3x + 6sin3x)/(2sin2⁡xcos2x) dx

= ∫(5cos3⁡x)/(2sin2xcos2x) dx + ∫(6sin3⁡x)/(2sin2⁡xcos2⁡x) dx

= 5/2 ∫(cos⁡x)/(sin2⁡x) dx + 3∫(sin⁡x)/(cos2⁡x) dx

= 5/2 ∫cot⁡xcosec⁡xdx + 3∫ sec⁡xtan⁡xdx

= (-5)/2 cosec⁡x + 3sec⁡x+c

= (-5)/2 cos⁡sec⁡x + 3sec⁡x+c

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