第 12 类 RD Sharma 解 - 第 19 章不定积分 - 练习 19.3 |设置 1

问题 1. 积分 ∫(2x – 3) ⁵ + √3x + 2 dx

解决方案：

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

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

解决方案：

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

问题 3. 整合$∫\frac{1}{(2-3x)} + \frac{1}{(\sqrt{3x-2})} dx$

解决方案：

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

问题 4. 整合$∫\frac{x+3}{(x+1)^4} dx$

解决方案：

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 

问题 5. 整合$∫\frac{1}{(\sqrt{x+1} + \sqrt{x})} dx$

解决方案：

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

问题 6. 整合$∫\frac{1}{(\sqrt{2x+3} + \sqrt{2x-3}} dx$

解决方案：

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

问题 7. 整合$∫\frac{2x}{(2x+1)^2} dx$

解决方案：

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

问题 8. 整合$∫\frac{1}{\sqrt{x+a} + \sqrt{x+b}} dx$

解决方案：

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

问题 9. 积分 ∫Sinx√1 + Cos2x dx

解决方案：

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 

问题 10. 整合$∫\frac{1+Cosx}{1-Cosx} dx$

解决方案：

Let I = $∫\frac{1+Cosx}{1-Cosx} dx$

= $∫\frac{2cos2(\frac{x}{2})}{2sin2(\frac{x}{2})} dx$

= ∫cot²(x/2) dx

= ∫cosec²(x/2) – 1 dx            -(1)

On integrating the equation(1), we get

= $\frac{-cot(\frac{x}{2})}{\frac{1}{2}} - x + c$

Hence, I = -2cot(x/2) – x + c