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

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 

= ∫3x^3/2dx + 4∫x^1/2 dx + 5∫dx

= x^(3/2)+1/(3/2 + 1) + 4x^(1/2)+1/(1/2 + 1) + 5x + c

= 6/5 x^5/2 + 8/3 x^3/2 + 5x + c

Question 2. Evaluate ∫(2^x + 5/x – 1/x^1/3)dx

Solution:

We have, ∫(2^x + 5/x – 1/x^1/3)dx

= ∫2^xdx + 5∫1/x dx – ∫1/x^1/3 dx

= 2^x/(log⁡2) + 5log⁡x – 3/2 x^2/3 + 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

= ∫ax^5/2 dx + ∫bx^3/2dx + ∫cx^1/2 dx

= (ax^(5/2)+1)/(5/2 + 1) + (bx^(3/2)+1)/(3/2 + 1) + (cx^(1/2)+1)/(1/2 + 1) + d

= (2ax^7/2)/7 + (2bx^5/2)/5 + (2cx^3/2)/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 + m^x + x^m + mx)dx

Solution:

We have, ∫(m/x + x/m + m^x + x^m + mx)dx

= m∫1/x dx + 1/m ∫xdx + ∫m^xdx + ⌋x^mdx + m∫xdx

= mlog⁡|x| + x²/2m + m^x/(log⁡m) + x^m+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 + ∫x^5/2 dx + 3∫x^3/2 dx + 3∫x^1/2 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

= x^1/2/(1/2) + x^7/2/(7/2) + (3x^5/2)/(5/2) + 3 x^3/2/(3/2) + c

= 2x^1/2 + 2/7 x^7/2 + 6/5 x^5/2 + 6/3 x^3/2 + c

= 2x^1/2 + 2/7 x^7/2 + 6/5 x^5/2 + 2x^3/2 + c

 Question 8. Evaluate ∫{x² + e^logx + (e/2)^x }dx

Solution:

We have, ∫{x² + e^logx + (e/2)^x }dx

= ∫x² dx + ∫e^logx 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 ∫ (x^e + e^x + e^e)dx

Solution:

We have, ∫ (x^e + e^x + e^e)dx                             

= ∫x^e dx + ∫e^xdx + ∫e^edx

= x^e+1/(e + 1) + e^x + e^ex + c

Question 10. Evaluate ∫√x (x³ – 2/x)dx 

Solution:

We have, ∫√x (x³ – 2/x)dx 

= ∫ x^7/2 dx – 2∫ x^-1/2 dx

= x^(7/2)+1/(7/2 + 1) – 2 x^(-1/2)+1/((-1)/2 + 1) + c

= x^9/2/(9/2) – (2x^-1/2)/((-1)/2) + c

= 2/9 x^9/2 – 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

= 2x^1/2 – 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)/x^1/3 + ∫x^1/2/x^1/3dx + ∫2/x^1/3dx

= ∫x^-2/3 dx + ∫x^1/6 dx + 2∫ x^-1/3dx

= 3x^1/3 + 6/7 x^7/6 + 3x^2/3 + c

Question 14. Evaluate ∫((1 + √x)²)/√x dx

Solution:

We have, ∫((1 + √x)²)/√x dx

= ∫(1 + x + 2√x)/x^1/2dx

= ∫x^-1/2+∫ x^1/2 dx + 2∫dx

= 2x^1/2 + 2/3 x^3/2 + 2x + c

= 2√x + 2x + 2/3 x^3/2 + c

Question 15. Evaluate ∫√x(3 – 5x)dx

Solution:

We have, ∫√x(3 – 5x)dx

= 3∫√x dx – 5x^3/2 dx

= 3x^3/2/(3/2) – 5 x^5/2/(5/2) + c

= 2x^3/2 – 2x^5/2 + c

Question 16. Evaluate ∫((x + 1)(x – 2))/√x dx

Solution:

We have, ∫((x + 1)(x – 2))/√x dx

= ∫(x² – 2x + x – 2)/x^1/2dx

= ∫(x² – x – 2)/x^1/2dx

= ∫x²/x^1/2dx – ∫x^1/2dx – 2∫x^-1/2dx

= (2x^5/2)/5 – (2x^3/2)/3 – 4x^1/2 + c

= 2/5 x^5/2 – (2x^3/2)/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