第12类RD Sharma解–第19章不定积分–练习19.10

问题1.评估^∫×2√x+ 2 DX

解决方案：

Let, I = ∫ x² √x + 2 dx (i)

Substituting x + 2 = t, x= t – 2

dx = dt

Substitute the above value in eq (i)

= ∫ (t – 2)² √t dt

= ∫ (t² + 4 – 4t) t^1/2 dt

= ∫(t^5/2 + 4t^1/2 – 4t^3/2) dt

Integrate the above eq then, we get

= t^7/2/(7/2) + 4t^3/2/(3/2) – 4t^5/2/(5/2) + c

=(2/7) t^7/2 + (8/3) t^3/2 – (8/5) t^5/2 + c

Now, put the value of t in above eq

=(2/7) (x+2)^7/2 + (8/3) (x+2)^3/2 – (8/5) (x+2)^5/2 + c

Hence, I =(2/7) (x+2)^7/2 + (8/3) (x+2)^3/2 – (8/5) (x+2)^5/2 + c

问题2.整合^{∫x 2} /(√x-1)dx

解决方案：

Let, I = ∫ x²/(√x-1) dx (i)

Put, x-1 = t, so the value of x=t+1

dx = dt

Put the above value in eq (i)

= ∫ (t+1)²/√t dt

On solving the above eq, we get

= ∫ (t² + 1 + 2t)/√t dt

= ∫ t^3/2 + t^-1/2 + 2t^-1/2 dt

Integrate the above eq then, we get

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

= (6t^5/2 + 30t^1/2 + 20t^3/2)/ 15 + c

= (2/15)t^1/2 (3t² + 15 + 10t) + c

= (2/15)(x -1)^1/2 (3(x -1)² + 15 + 10(x -1)) + c

= (2/15)(x -1)^1/2 (3(x² + 1 – 2x) + 15 + 10x -10)) + c

= (2/15)(x -1)^1/2 (3x² + 3 – 6x + 15 + 10x -10)) + c

= (2/15)(√x -1) (3x² + 4x + 8) + c

Hence, I = = (2/15)(√x -1) (3x² + 4x + 8) + c

问题3.整合^{∫x 2} /(√3x+ 4)dx

解决方案：

Let, I = ∫ x²/(√3x+4) dx (i)

Put, 3x + 4 = t, so the value of x = (t – 4)/3

dx = dt/3

Put the above value in eq (i)

= ∫ ((t – 4)/3)²/ √t dt/3

= (1/3) ∫ (t² + 16 – 8t)/ 9√t dt

= (1/27) ∫ (t² + 16 – 8t)/√t dt

= (1/27) ∫ (t^3/2 + 16t^-1/2 – 8t^1/2) dt

Integrate the above eq then, we get

= (1/27) [(2/5)t^5/2 – (16/3)t^3/2 + 32t^1/2]+ c

Now put the value of t in above eq

= (1/27) [(2/5)(3x + 4)^5/2 – (16/3)(3x + 4)^3/2 + 32(3x + 4)^1/2]+ c

Hence, I =(1/27) [(2/5)(3x + 4)^5/2 – (16/3)(3x + 4)^3/2 + 32(3x + 4)^1/2]+ c

问题4.整合∫(2x-1)/(x-1) ² dx

解决方案：

Let, I = ∫ (2x-1)/ (x-1)² dx

Substituting x-1 = t and dx = dt, we get

= ∫ 2(t + 1)-1 / t² dt

= ∫ (2t + 2 – 1)/ t² dt

= ∫ (2t +1) / t² dt

= ∫ 2t/t² +1/ t² dt

= 2∫ 1/t dt + ∫ t^-2 dt

Integrate the above eq then, we get

= 2log |t| – t^-1 + c

Put the value of t in above eq

= 2log |x-1| – 1/(x-1) + c

Hence, I = 2log |x-1| – 1/(x-1) + c

问题5.整合∫(2x ² + 3)√x+2 dx

解决方案：

Let, I = ∫(2x² + 3) √x +2 dx

Substituting x +2 = t and dx = dt, we get

= ∫ [2(t -2)² + 3] √t dt

= ∫ [2(t² + 4 – 4t) + 3] √t dt

= ∫ [2t² + 8 – 8t + 3] √t dt

= ∫ [2t^5/2 + 11t^-1/2 – 8t^3/2] dt

Integrate the above eq then, we get

= (4/7)t^7/2 + (22/3)t^3/2 – (16/5) t^5/2 + c

= (4/7)(x+2)^7/2 + (22/3)(x+2)^3/2 – (16/5)(x+2)^5/2 + c

Hence, I = (4/7)(x+2)^7/2 + (22/3)(x+2)^3/2 – (16/5)(x+2)^5/2 + c

问题6.整合∫(x ² + 3x + 1)/(x + 1) ² dx

解决方案：

Let, I = ∫ (x² + 3x + 1)/ (x+1)² dx

Substituting x + 1 = t and dx = dt, we get

= ∫ [(t – 1)² + 3(t – 1) + 1]/ t² dt

= ∫ (t² + 1 – 2t +3t -3 +1)/ t² dt

= ∫ (t² + t – 1)/ t² dt

= ∫ t²/t² + t/ t² – 1/t² dt

= ∫ (1 + 1/t – t^-2) dt

Integrate the above eq then, we get

= t + log |t| + 1/t + c

Put the value of t in above eq

= (x +1) + log |x +1| + 1/(x+1) + c

问题7.对^{∫x 2} /(√1-x)dx进行积分

解决方案：

Let, I =∫ x² / (√1-x) dx

Substituting 1- x = t and dx = – dt, we get

= ∫ – (1-t)² /√t dt

= – ∫(1 + t² – 2t)/ √t dt

= – ∫ t^-1/2 + t^3/2 – 2t^1/2 dt

Integrate the above eq then, we get

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

= – (30t^1/2 + 6t^5/2 – 20t^3/2) / 15 + c

= – 2t^1/2 /15(15 + 3t² – 10t) + c

= (- 2/15) √(1-x) (15 + 3(1 -x)² – 10(1 -x)) + c

= (- 2/15) √(1-x) (15 + 3(1 + x² – 2x) – 10 + 10x)) + c

= (- 2/15) √(1-x) (5 + 3 + 3x² – 6x + 10x) + c

= (- 2/15) √(1-x) (3x² + 4x + 8) + c

Hence, I = (-2/15) (3x² + 4x + 5) √1-x + c

问题8.整合∫x(1 – x) ²³ dx

解决方案：

Let, I = ∫ x(1 – x)²³ dx

Substituting 1- x=t and dx = -dt, we get

= – ∫(1-t)t²³ dt

= – ∫ (t²³ – t²⁴) dt

= ∫ (t²⁴ – t²³) dt

Integrate the above eq then, we get

= t²⁵/25 – t²⁴/24 + c

= (1-x)²⁵/25 – (1-x)²⁴/24 + c

Hence, I = (1-x)²⁵/25 – (1-x)²⁴/24 + c