Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.3 | Set 1

Question 1. Integrate ∫(2x - 3)⁵  + √3x + 2 dx

Solution:

Let I = ∫(2x - 3)⁵ + √3x + 2 dx         -(1)

On integrating the equation(1), we get

= \frac{(2x - 3)^6}{(2 × 6)} + \frac{(3x + 2)^{\frac{3}{2}}}{3×\frac{3}{2}} + c

= \frac{(2x - 3)^6}{12} + \frac{2(3x + 2)^{\frac{3}{2}}}{9} + c

Therefore, I = \frac{(2x - 3)^6}{12} + \frac{2(3x + 2)^{\frac{3}{2}}}{9} + c

Question 2. Integrate ∫\frac{1}{(7x-5)^3} + \frac{1}{\sqrt{5x-4}}dx

Solution:

Let I =∫\frac{1}{(7x-5)^3} + \frac{1}{\sqrt{5x-4}}dx   

I = ∫(7x - 5)^-3 + (5x - 4)^-1/2 dx            -(1)

On integrating the equation(1), we get

= \frac{(7x - 5)^{-2}}{7 × (-2)} + \frac{(5x - 4)^{\frac{1}{2}}}{(5 × \frac{1}{2})} + c

= \frac{-(7x - 5)^{-2}}{14} + \frac{2}{5}\sqrt{(5x - 4)} + c

Hence, I = \frac{-(7x - 5)^{-2}}{14} + \frac{2}{5}\sqrt{(5x - 4)} + c

Question 3. Integrate ∫\frac{1}{(2-3x)} + \frac{1}{(\sqrt{3x-2})} dx

Solution:

Let I = ∫\frac{1}{(2-3x)} + \frac{1}{(\sqrt{3x-2})} dx

= ∫1/(2 - 3x) + (3x - 2)^-1/2 dx            -(1)

On integrating the equation(1), we get

= log |2 - 3x| /(-3) + (2/3) × (3x - 2)^1/2 + c

= (-1/3) log|2 - 3x| + (2/3)√3x - 2 + c

Hence, I = (-1/3) log|2 - 3x| + (2/3)√3x - 2 + c

Question 4. Integrate ∫\frac{x+3}{(x+1)^4} dx

Solution:

Let I = ∫\frac{x+3}{(x+1)^4} dx

= ∫\frac{(x+1+2)}{(x+1)^4} dx

= ∫\frac{(x+1)}{(x+1)^4} dx + 2∫\frac{1}{(x+1)^4} dx

= ∫\frac{1}{(x+1)^3} dx + 2∫(x+1)-4 dx

= ∫(x + 1)^-3 dx + 2∫(x + 1)^-4 dx            -(1)

On integrating the equation(1), we get

= (x + 1)^-2/(-2) + 2×(x + 1)^-3/(-3) + c

= (-1/2)(x + 1)^-2 + (-2/3)(x + 1)^-3 + c

Hence, I = (-1/2)(x + 1)^-2 + (-2/3)(x + 1)^-3 + c 

Question 5. Integrate ∫\frac{1}{(\sqrt{x+1} + \sqrt{x})}  dx

Solution:

Let I = ∫\frac{1}{(\sqrt{x+1} + \sqrt{x})}  dx

= ∫\frac{\frac{1}{(\sqrt{x+1} + \sqrt{x})} × \frac{1}{(\sqrt{x+1} - \sqrt{x})}} {\frac{1}{(\sqrt{x+1} - \sqrt{x})} }dx

= ∫\frac{(\sqrt{x+1} - \sqrt{x}) }{ (\sqrt{x+1})^2 -(\sqrt{x})^2} dx

⁼ ∫\frac{(\sqrt{x+1} - \sqrt{x})}{x+1-x}  dx

= ∫(√x + 1 - √x) dx

= ∫(x + 1)^1/2 - (x)^1/2 dx            -(1)

On integrating the equation(1), we get

= \frac{(x + 1)^{\frac{3}{2}}}{\frac{3}{2}}  - \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c

= (2/3)(x + 1)^3/2  - (2/3)(x)^3/2 + c

Hence, I = (2/3)(x + 1)^3/2 - (2/3)(x)^3/2 + c

Question 6. Integrate ∫\frac{1}{(\sqrt{2x+3} + \sqrt{2x-3}} dx

Solution: 

Let I = ∫\frac{1}{(\sqrt{2x+3} + \sqrt{2x-3}} dx

Now multiply with the conjugate

= ∫\frac{1}{(\sqrt{2x+3} + \sqrt{2x-3})} × \frac{(\sqrt{2x+3} - \sqrt{2x-3})}{(\sqrt{2x+3} - \sqrt{2x-3})}dx

= ∫\frac{(\sqrt{2x+3} - \sqrt{2x-3})}{(\sqrt{2x+3})^2 - (\sqrt{2x-3})^2} dx

= ∫\frac{(\sqrt{2x+3} - \sqrt{2x-3}}{(2x+3-2x+3)} dx

= ∫\frac{\sqrt{2x+3} - \sqrt{2x-3})}{6}  dx

= 1/6 ∫(√2x + 3 - √2x - 3) dx

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

On integrating the equation(1), we get

= \frac{1}{6} × \frac{(2x + 3)^{\frac{3}{2}}}{\frac{3}{2} × 2} - \frac{1}{6}×\frac{(2x - 3)^{\frac{3}{2}}}{\frac{3}{2} × 2} + c

= \frac{1}{18} × (2x + 3)^{\frac{3}{2}} - \frac{1}{18}(2x - 3)^{\frac{3}{2}} + c

Hence, I = \frac{1}{18} × (2x + 3)^{\frac{3}{2}} - \frac{1}{18}(2x - 3)^{\frac{3}{2}} + c

Question 7. Integrate ∫\frac{2x}{(2x+1)^2} dx

Solution:

Let I = ∫\frac{2x}{(2x+1)^2} dx

= ∫\frac{2x+1-1}{(2x+1)^2} dx

= ∫\frac{2x+1}{(2x+1)^2} dx - \frac{1}{(2x+1)^2} dx

= ∫\frac{1}{(2x+1)} dx  - \frac{1}{(2x+1)^2} dx

= ∫1/(2x + 1) dx - (2x + 1)^-2 dx            -(1)

On integrating the equation(1), we get

= log|2x + 1| × \frac{1}{2} - \frac{(2x+1)^{-1}}{\frac{-1}{2}} + c

= (1/2) log|2x + 1| + (1/2)(2x + 1)^-1 + c

Hence, I= (1/2) log|2x + 1| + (1/2)(2x + 1)^-1 + c

Question 8. Integrate ∫\frac{1}{\sqrt{x+a} + \sqrt{x+b}} dx

Solution:

Let I = ∫\frac{1}{\sqrt{x+a} + \sqrt{x+b}} dx

Now multiply with the conjugate,

= ∫\frac{1}{\sqrt{x+a} + \sqrt{x+b}} × \frac{(\sqrt{x+a} - \sqrt{x+b})}{(\sqrt{x+a} - \sqrt{x+b})} dx

= ∫\frac{(\sqrt{x+a} - \sqrt{x+b})}{(\sqrt{x+a})^2- (\sqrt{x+b})^2} dx

= ∫\frac{(\sqrt{x+a} - \sqrt{x+b})}{(x+a-x-b)} dx

= (1/a - b)∫(x + a)^1/2 - (x + b)^1/2 dx            -(1)

On integrating the equation(1), we get

= (\frac{1}{a - b}) (\frac{(x+a)^{3/2}}{\frac{3}{2}} - \frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}})  +c

= (1/a - b) ((2/3) (x + a)^3/2 - (2/3) (x + b)^3/2) + c

Hence, I = (1/a - b) (2/3) ((x + a)^3/2 - (x + b)^3/2) + c

Question 9. Integrate ∫Sinx√1 + Cos2x dx

Solution:

Let I = ∫Sinx√1 + Cos2x dx

On  substituting the formula, we get

= ∫Sinx√(2Cos²x) dx

= ∫Sinx√2Cosx dx

= √2 ∫Sinx Cosx dx

Multiply and divide the above equation by 2

= √2/2∫2SinxCosx dx

= √2/2∫Sin2x dx            -(1)

On integrating the equation(1), we get

= √2/2 (-Cos2x/2) + c

Hence, I = (-1/2√2) Cos2x + c 

Summary

Exercise 19.3 Set 1 in Chapter 19 on Indefinite Integrals focuses on integrating rational functions where the degree of the numerator is greater than the degree of the denominator, and the denominator is a linear factor. This set of problems requires students to apply polynomial long division before integration. After division, students will typically encounter a polynomial term plus a constant divided by the linear factor. The polynomial term is integrated using standard integration rules, while the fraction term leads to a logarithmic function. This exercise aims to enhance students' skills in handling rational functions, reinforcing their understanding of polynomial division and integration techniques for different types of functions, including logarithmic integrals.