问题1.评估:∫(x 2 + x + 1)/(x 2 – x)dx
解决方案:
Given that I = ∫(x2 + x + 1)/(x2 – x) dx
= ∫ [1 + (2x + 1)/(x2 – x)]dx
= x + ∫(2x + 1)/(x2 – x) dx + c1
= x + I1 + c1 …….(i)
Now, I1 = ∫(2x + 1)/(x2 – x) dx
Let 2x + 1 = λ d/dx (x2 – x) + μ = λ(2x – 1) + μ
2x + 1 = (2λ)x – λ + μ
By comparing the coefficients of x, we get
2 = 2λ ⇒ λ = 1
-λ + μ = 1 ⇒ μ = 2
I1 = ∫ ((2x – 1) + 2)/(x2 – x) dx)
= ∫(2x – 1)/(x2 – x) dx + 2∫1/(x2 – x) dx
= ∫(2x – 1)/(x2 – x) dx + 2∫1/(x2 – 2x(1/2) + (1/2)2 – (1/2)2) dx
= ∫(2x – 1)/(x2 – x) dx + 2∫1/((x – 1/2)2 – (1/2)2) dx
= log|x2 – x| +
As we know that ∫1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
So, I1 = log|x2 – x| + 2log|(x – 1)/x| + c2 ……(ii)
Now put the value of I1 in eq(i), we get
I = x + log|x2 – x| + 2log|(x – 1)/x| + c
问题2。∫(x 2 + x – 1)/(x 2 + x – 6)dx
解决方案:
Given that I = ∫ (x2 + x – 1)/(x2 + x – 6) dx
= ∫[1 + 5 /(x2 + x – 6)]dx
I = x + ∫5/(x2 + x – 6) dx + c1
Let us assume I1 = 5∫1/(x2 + x – 6) dx
I = x + I1 + c1 …..(i)
= 5∫ 1/(x2 + 2x(1/2) + (1/2)2 – (1/2)2 – 6) dx
= 5∫ 1/((x + 1/2)2 – (5/2)2dx
= 5 1/2(5/2) log|(x + 1/2 – 5/2)/(x + 1/2 + 5/2)| + c2
As we know that ∫ 1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
So, we get
I1 = log|(x – 2)/(x + 3)| + c2 …….(ii)
Now put the value of I1 in eq(i), we get
I = x + log|(x – 2)/(x + 3)| + c
问题3。∫(1 – x 2 )/(x(1 – 2x))dx
解决方案:
Given that I = ∫ (1 – x2)/(x(1 – 2x)) dx
= ∫ (1 – x2)/(x – 2x2) dx
= ∫ (x2 – 1)/(2x2 – x) dx
= ∫ [1/2 + (x/2 – 1)/(2x2 – x)]dx
I = 1/2x + ∫(x/2 – 1)/(2x2 – x) dx + c1
Let us assume I1 = ∫(x/2 – 1)/(2x2 – x) dx
So, I = 1/2x + I1 + c1 …..(i)
Now, let x/2 -1 = λ d/dx (2x2 – x) + μ = λ(4x – 1) + μ
x/2 – 1 = (4λ)x – λ + μ
By comparing the coefficients of x, we get
1/2 = 4λ ⇒ λ = 1/8
-λ + μ = -1 ⇒ -(1/8) + μ = -1
μ = -7/8
I1 = ∫ (1/8(4x – 1) – 7/8)/(2x2 – x) dx
= 1/8 ∫(4x – 1)/(2x2 – x) dx – 7/8 ∫1/2(x2 – x/2)dx
= 1/8 ∫(4x – 1)/(2x2 – x) dx – 7/16 ∫1/(x2 – 2x(1/4) + (1/4)2 – (1/4)2) dx
= 1/8 ∫(4x – 1)/(2x2 – x) dx – 7/16 [1/((x – 1/4)2 – (1/4)2) dx
= 1/8 log|2x2 – x| – 7/16 × 1/2(1/4) log|(x – 1/4 – 1/4)/(x – 1/4 + 1/4)| + c2
As we know that ∫ 1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
So, we get
I1 = 1/8 log|x| + 1/8 log|2x – 1| – 7/8 log|1 – 2x| + 7/8 log2 + 7/8 log|x| + c2
I1 = log|x| – 3/4 log|1 – 2x| + c3 [Here, c3 = c2 + 7/8 log2]
Now put the value of I1 in eq(i), we get
I = 1/2x + log|x| – 3/4 log|1 – 2x| + c
问题4.∫(x 2 +1)/(x 2 – 5x + 6)dx
解决方案:
Given that I = ∫(x2 + 1)/(x2 – 5x + 6)dx
Now we convert I into proper rational function by dividing x2 + 1 by x2 – 5x + 6
So,
(x2 + 1)/(x2 – 5x + 6) = 1 + (5x – 5)/(x2 – 5x + 6) = 1 + (5x – 5)/((x – 2)(x – 3))
Let
(5x – 5)/((x – 2)(x – 3)) = A/(x – 2) + B/(x – 3)
So, we get A + B = 5 and 3A + 2B = 5
On solving both the equations we get A = -5 and B = 10
So,
Hence, ∫(x2 + 1)/((x + 1)2 (x + 3)) dx =∫dx – 5∫1/(x – 2) dx + 10∫x/(x – 3)
I = x – 5log|x – 2| + 10log|x – 3| + c
问题5.∫x2 /(X 2 + 7×+ 10)DX
解决方案:
Given that I = ∫x2/(x2 + 7x + 10) dx
= ∫ {1 – (7x + 10)/(x2 + 7x + 10)}dx
I = x – ∫(7x + 10)/(x2 + 7x + 10) dx + c1
Let us assume I1 = ∫(7x + 10)/(x2 + 7x + 10) dx
So, I = x – I1 + c1 …..(i)
Now let us assume 7x + 10 = λ d/dx (x2 + 7x + 10) + μ = λ(2x + 7) + μ
7x + 10 = (2λ)x + 7λ + μ
By comparing the coefficients of x, we get
7 = 2λ ⇒ λ = 7/2
7λ + μ = 10 ⇒ 7(7/2) + μ = 10μ = -29/2
So, I1 = ∫(7/2(2x + 7) – 29/2)/(x2 + 7x + 10) dx
= 7/2 ∫((2x + 7))/(x2 + 7x + 10) dx – 29/2∫1/(x2 + 2x(7/2) + (7/2)2 – (7/2)2 + 10) dx
= 7/2 ∫(2x + 7)/(x2 + 7x + 10) dx – 29/2 {1/((x + 7/2)2 – (3/2)2) dx
= 7/2 log|x2 + 7x + 10| – 29/2 × 1/2(3/2) log|(x + 7/2 – 3/2)/(x + 7/2 + 3/2)| + c2
As we know that ∫ 1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
So, we get
I1 = 7/2 log|x2 + 7x + 10| – 29/6 log|(x + 2)/(x + 5)| + c2
Now put the value of I1 in eq(i), we get
I = x – 7/2 log|x2 + 7x + 10| + 29/6 log|(x + 2)/(x + 5)| + c
问题6.∫(x 2 + x + 1)/(x 2 – x + 1)dx
解决方案:
Given that l = ∫(x2 + x + 1)/(x2 – x + 1) dx
= ∫[1 + 2x/(x2 – x + 1)]dx
= x + ∫2x/(x2 – x + 1) dx + c1
Let us assume I1 = ∫2x/(x2 – x + 1) dx
So, I = x + I1 + c1 …..(i)
Now let 2x = λ d/dx (x2 – x + 1) + μ = λ(2x – 1) + μ
2x = (2λ) × -λ + μ
By comparing the coefficients of x, we get
2 = 2λ ⇒ λ = 1
-λ + μ = 0 ⇒ -1 + μ = 0
μ = 1
So, I1 = ∫((2x – 1) + 1)/(x2 – x + 1) dx
= ∫((2x – 1))/(x2 – x + 1) dx + ∫1/(x2 – 2x(1/2) + (1/2)2 – (1/2)2 + 1) dx
= ∫(2x – 1)/(x2 – x + 1) dx + ∫1/((x – 1/2)2 + (√3/2)2) dx
= log|x2 – x + 1| + 2/√3 tan-1((x – 1/2)/(√3/2)) + c2
As we know that, ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c
So, we get
I1 = log|x2 – x + 1| + 2/√3 tan-1((2x – 1)/√3) + c2
Now put the value of I1 in eq(i), we get
I = x + log|x2 – x + 1| + 2/√3 tan-1((2x – 1)/√3) + c
问题7.∫(x – 1) 2 /(x 2 + 2x + 2)dx
解决方案:
Given that, I = ∫(x – 1)2/(x2 + 2x + 2) dx
= ∫(x2 – 2x + 1)/(x2 + 2x + 2) dx
= ∫[1 – (4x + 1)/(x2 + 2x + 2)]dx
= x – ∫(4x + 1)/(x2 + 2x + 2) dx + c1
Let us assume l1 = ∫(4x + 1)/(x2 + 2x + 2) dx
So, I = x – I1 + c1 …..(i)
Now, let 4x + 1 = λ d/dx (x2 + 2x + 2) + μ
= λ(2x + 2) + μ = (2k)x + (2λ + μ)
By comparing the coefficients of x, we get
4 = 2λ ⇒ λ = 2
2λ + μ = 1 ⇒ 2(2) + μ = 1
μ = -3
l1 = ∫ (2(2x + 2) – 3)/(x2 + 2x + 2) dx
= 2∫ ((2x + 2))/(x2 + 2x + 2) dx – 3∫1/(x2 – 2x + (1)2 – (1)2 + 2) dx
= 2∫ (2x + 2)/(x2 + 2x + 2) dx – 3∫1/((x + 1)2 + (1)2) dx
As we know that, ∫1/(x2 + 1) dx = tan-1x + c
So, we get
l1 = 2log|x2 + 2x + 2| – 3tan-1(x + 1) + c2
Now put the value of I1 in eq(i), we get
I = x – 2log|x2 + 2x + 2| + 3tan-1(x + 1) + c
问题8.∫(x 3 + x 2 + 2x +1)/(x 2 – x + 1)dx
解决方案:
Given that, I = ∫(x3 + x2 + 2x + 1)/(x2 – x + 1) dx
= ∫[x + 2 + (3x – 1)/(x2 – x + 1)]dx
= x2/2 + 2x + ∫(3x. -1)/(x2 – x + 1) dx + c1
Let us assume l1 = ∫(3x – 1)/(x2 – x + 1) dx
So, I = x2/2 + 2x + I1 + c1 …..(i)
Now, let 3x – 1 = λ d/dx (x2 – x + 1) + μ = λ(2x – 1) + μ
3x – 1 = (2λ)x – λ + μ
By comparing the coefficients of x, we get
3 = 2λ ⇒ λ = 3/2
-λ + μ = -1 ⇒ -(3/2) + μ = -1
μ = 1/2
So, I1 = ∫(3/2(2x – 1) + 1/2)/(x2 – x + 1) dx
= 3/2 ∫ ((2x – 1))/(x2 – x + 1) dx + 1/2 ∫1/(x2 – 2x(1/2) + (1/2)2 – (1/2)2 + 1) dx
= 3/2 ∫ (2x – 1)/(x2 – x + 1) dx + 1/2 ∫1/((x + 1/2)2 + (√3/2)2) dx
= 3/2 log|x2 – x + 1| + 1/2 × 2/√3 tan-1((x + 1/2)/(√3/2)) + c2
As we know that, ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c
So, we get
I1 = 3/2 log|x2 – x + 1| + 1/√3 tan-1((2x + 1)/√3) + c2
Now put the value of I1 in eq(i), we get
I = x2/2 + 2x + 3/2 log|x2 – x + 1| + 1/√3 tan-1((2x + 1)/√3) + c
问题9.∫(x 2 (x 4 + 4))/(x 2 + 4)dx
解决方案:
Given that, I = ∫(x2 (x4 + 4))/((x2 + 4)) dx
= ∫ (x6 + 4x2)/((x2 + 4)) dx
= ∫ [x4 – 4x2 + 20 – 80/(x2 + 4)]dx
= x5/5 – (4x3)/3 + 20x – 80 ∫1/(x2 + 4) dx + c1
Let us assume I1 = ∫1/(x2 + 4) dx
So, I = x5/5 – (4x3)/3 + 20x – 80I1 + c1 …..(i)
Now, I1 = ∫1/(x2 + (2)2) dx
As we know that, ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c
So, we get
I1 = 1/2 tan-1(x/2) + c2
Now put the value of I1 in eq(i), we get
I = x5/5 – (4x3)/3 + 20x – 80/2 tan-1(x/2) + c
I = x5/5 – (4x3)/3 + 20x – 40tan-1(x/2) + c
问题10.∫x 2 /(x 2 + 6x + 12)dx
解决方案:
Given that, l = ∫ x2/(x2 + 6x + 12) dx
= ∫ [1 – (6x + 12)/(x2 + 6x + 12)]dx
= x – ∫(6x + 12)/(x2 + 6x + 12) dx + c1
Let us assume I1 = ∫(6x + 12)/(x2 + 6x + 12) dx
So, I = x – I1 + c1 …..(i)
Now, let 6x + 12 = λ d/dx (x2 + 6x + 12) + μ = λ(2x + 6) + μ
6x + 12 = (2λ)x + 6λ + μ
By comparing the coefficients of the power of x, we get
6 = 2λ ⇒ λ = 3
6λ + μ = 12
6(3) + μ = 12
μ = -6
So, l1 = ∫(3(2x + 6) – 6)/(x2 + 6x + 12) dx
=3∫ ((2x + 6))/(x2 + 6x + 12) dx – 6∫1/(x2 + 2x(3) + (3)2 – (3)2 + 12) dx
= 3∫ (2x + 6)/(x2 + 6x + 12) dx + 6∫1/((x + 3)2 + (√3)2) dx
As we know that, ∫1/(x2 + a2) dx = 1/2 tan-1(x/a) + c
So, we get
I1 = 3log|x2 + 6x + 12| + 6/√3 tan-1((x + 3)/√3) + c2
I1 = 3log|x2 + 6x + 12| + 2√3 tan-1((x + 3)/√3) + c2
Now put the value of I1 in eq(i), we get
l = x – 3log|x2 + 6x + 12| + 2√3 tan-1((x + 3)/√3) + c