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

Summary

Exercise 19.2 Set 1 in Chapter 19 on Indefinite Integrals focuses on integrating rational functions where the numerator is linear (ax + b) and the denominator is a difference of squares (x^2 - a^2). This set of problems requires students to use partial fraction decomposition, splitting the fraction into two simpler fractions. The resulting fractions typically lead to logarithmic functions. This exercise aims to enhance students' skills in handling rational functions with factorable quadratic denominators, reinforcing their understanding of partial fraction decomposition and integration techniques for logarithmic functions. The problems in this set are designed to help students recognize the pattern of difference of squares and apply the appropriate integration method efficiently.