问题1.∫x /(x 2 + 3x + 2)dx
解决方案:
Given that I = ∫ x/(x2 + 3x + 2) dx
Let x = m d/dx (x2 + 3x + 2) + n
= m(2x + 3) + n
x = (2m)x + (3λ + n)
On comparing the coefficients of x,
2m = 1
m = 1/2
3m + n = 0
3(1/2) + n = 0
n = -3/2
I = ∫(1/2(2x + 3) – 3/2)/(x2 + 3x + 2) dx
= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx – 3/2 ∫1/(x2 + 3x + 2) dx
= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx – 3/2 ∫1/(x2 + 2x(3/2) + (3/2)2 – (3/2)2 + 2) dx
= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx – 3/2 ∫1/((x + 3/2)2 – (1/2)2) dx
= 1/2log|x2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 3/2 – 1/2)/(x + 3/2 + 1/2)| + c
As we know that ∫1/(a2 – x2)dx = 1/2a log|(x – a)/(x + a)| + c]
Hence, I = 1/2log|x2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 1)/(x + 2)| + c
问题2.∫(x +1)/(x 2 + x + 3)dx
解决方案:
Given that I = ∫(x + 1)/(x2 + x + 3) dx
Let us considered x + 1 = m d/dx(x2 + x + 3) + n
x + 1 = m(2x + 1) + n
x + 1 = (2m)x + (m + n)
On comparing the co-efficient of x,
2m = 1
m = 1/2
m + n = 1
(1/2) + n = 1
n = 1/2
Now,
I = ∫(1/2(2x + 1) + 1/2)/(x2 + x + 3) dx
=1/2∫(2x + 1)/(x2 + x + 1) dx + 1/2∫1/(x2 + 2x(1/2) + (1/2)²-(1/2)²+3) dx
= 1/2∫(2x + 1)/(x2 + x + 1) dx + 1/2∫1/((x + 1/2)2 + (11/4)) dx
= 1/2 ∫(2x + 1)/(x2 + x + 1) dx + 1/2 ∫1/((x + 1/2)2 + (√11/2)2) dx
= 1/2 log|x2 + x + 3| + (1/2) * (1/(√11/2)) tan-1((x + 1/2)/(√11/2)) + c
As we know that ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c
Hence, I = 1/2 log|x2 + x + 3| + 1/√11 tan-1((2x + 1)/(√11)) + c
问题3。∫(x – 3)/(x 2 + 2x – 4)dx
解决方案:
Given that I = ∫(x – 3)/(x2 + 2x – 4) dx
Let us considered x – 3 = m d/dx (x2 + 2x – 4) + n
= m(2x + 2) + n
x – 3 = (2m)x + (2m + n)
On comparing the coefficients of x,
2m = 1
m = 1/2
2m + n = -3
2(1/2) + n = -3
n = -4
So,
I = ∫(1/2(2x + 2) – 4)/(x2 + 2x – 4) dx
= 1/2 ∫(2x + 2)/(x2 + 2x – 4) dx – 4∫ 1/(x2 + 2x + (1)2 – (1)2 – 4) dx
= 1/2 ∫(2x + 2)/(x2 + 2x – 4) dx – 4∫ 1/((x + 1)2 – (√5)) dx
As we know that ∫1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
= 1/2 log|x2 + 2x – 4| – 4 × 1/(2√5) log|(x + 1 – √5)/(x + 1 + √5)| + c
Hence, I = 1/2 log|x2 + 2x – 4| – 2/√5 log|(x + 1 – √5)/(x + 1 + √5)| + c
问题4.∫(2x – 3)/(x 2 + 6x + 13)dx
解决方案:
Given that I = ∫ (2x – 3)/(x2 + 6x + 13) dx
Let us considered 2x – 3 = m d/dx (x2 + 6x + 13) + n
= m(2x + 6) + n
2x – 3 = (2m)x + (6m + n)
On comparing the co-efficient of x, we get
2m = 2
m = 1
6m + n = -3
6 * 1 + n = -3
n = -9
Now,
= ∫(1 * (2x + 6) – 9)/(x2 + 6x + 13) dx
= ∫(2x + 6)/(x2 + 6x + 13) dx + ∫(-9)/(x2 + 2 * (3) * x + (3)2 – (3)2 + 13) dx
= ∫(2x + 6)/(x2 + 6x + 13) dx -9 ∫1/((x + 3)2 + (2)) dx
= log|(x2 + 6x + 13)| – 9 * (1/2) tan-1((x + 3)/2) + c
Hence, I = log|(x2 + 6x + 13)| – 9 × (1/2) tan-1((x + 3)/2) + 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 ……..(i)
Let I1 = ∫(7x + 10)/(x2 + 7x + 10) dx
Let 7x + 10 = md/dx (x2 + 7x + 10) + n
= m(2x + 7) + n
7x + 10 = (2m)x + 7m + n
On comparing the coefficients of like powers of x,
7 = 2m
m = 7/2
7m + n = 10
7(7/2) + n = 10
n = -29/2
So, l = ∫(1/6(6x – 4) – 1/3)/(3x2 – 4x + 3) dx
= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x2 – 4/3x + 1) dx
= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x2 – 2x(2/3) + (2/3)2 – (2/3)2 + (2)2) dx
= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x – 2/3)2 + (√5/2)) dx
= 1/6log|(3x2 – 4x + 3) | – ((1/9) × 1/(√5/3)) tan((x – 2/3)/(√5/3)) + c
Hence, I = 1/6log|(3x2 – 4x + 3)| – (√5/15)tan-1((3x – 2)/√5) + c
问题6.∫2x/(2 + x – x 2 )dx
解决方案:
Given that I = ∫2x/(2 + x – x2) dx
Now,
2x = m(d/dx(2 + x + x2)) + n
2x = m(-2x + 1) + n
Now equating the co-efficient of we will get m, n
m = -1,
n = 1
∫2x/(2 + x – x2) dx
= ∫(m(-2x + 1) + n)/(2 + x – x2) dx
= ∫(-1(-2x + 1) + 1)/(2 + x – x2) dx
= ∫(-1(-2x + 1))/(2 + x – x2) dx + 1/(2 + x – x2) dx
= -log|2 + x – x2| + ∫1/(2 + x – x2) dx
= -log|2 + x – x2| – ∫1/(x2 – x – 2) dx
= -log|2 + x – x2 – ∫1/(x2 – x(1/2)(2) + (1/2) – (1/2) – 2) dx
= -log|2 + x – x2| + ∫1/((x – 1/2)2 – (3/2)2) dx
= -log|2 + x – x2| – 1/3 log|((x – 1/2) – (3/2))/((x – 1/2) + (3/2))| + c
Hence, I = -log|2 + x – x2| – 1/3 log|(x – 2)/(x + 1)| + c
问题7.∫(1 – 3x)/(3x 2 + 4x + 2)dx
解决方案:
Given that I = ∫(1 – 3x)/(3x2 + 4x + 2) dx
Let us considered 1 – 3x = m d/dx (3x2 + 4x + 2) + n
= m(6x + 4) + n – 11 – 3x = (6m)× + (4λ + n)
On comparing the coefficients of l x,
6m = -3
m = -1/2
4m + n = 1
4(-1/2) + n = 1
n = 3
I = ∫(-1/2(6x + 4) + 3)/(3x2 + 4x + 2) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + 3∫1/(3x2 + 4x + 2) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + 3/3 ∫1/(x2 + 4/3x + 2/3) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/(x2 + 2x(2/3) + (2/3)2 – (2/3)2 + (2/3) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + 2/9) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + (√2/3)2) dx
= -1/2 log|(3x2 + 4x + 2) | + 3/√2tan-1((x + 2/3)/(√2/3)) + c
Hence, I = -1/2 log|(3x2 + 4x + 2)| + 3/√2tan-1((3x + 2)/√2) + c
问题8.∫(2x + 5)/(x 2 – x – 2)dx
解决方案:
Given that I = ∫(2x + 5)/(x2 – x – 2) dx
Let 2x + 5 = md/dx (x2 – x – 2) + n
= m(2x – 1) + n
2x + 5 = (2m)x – m + n
On comparing the coefficients of x,
2m = 2
m = 1
-m + n = 5
-1 + n = 5
n = 6
So,
I = ∫((2x – 1) + 6)/(x2 – x – 2) dx
= ∫ ((2x – 1))/(x2 – x – 2) dx + 6∫1/(x2 – 2x(1/2) + (1/2)2 – (1/2)2 – 2) dx
= ∫ (2x – 1)/(x2 – x – 2) dx + 6∫1/((x – 1/2)2 – 9/4) dx
= ∫ (2x – 1)/(x2 – x – 2) dx + 6∫1/((x – 1/2)2 – (3/2)2) dx
= log|x2 – x – 2| + 6/2(3/2) log|(x – 1/2 – 3/2)/(x – 1/2 + 3/2)| + c
As we know that ∫1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
Hence, I = log|x2 – x – 2| + 2log|(x – 2)/(x + 1)| + c
问题9.∫(ax 3 + bx)/(x 4 + c 2 )dx
解决方案:
Given that I = ∫(ax3 + bx)/(x4 + c2) dx
Let us considered ax3 + bx = md/dx (x4 + c2) + n
ax3 + bx = n(4x3) + n
On comparing the coefficients of x,
4m = a
m = a/4
and
n = 0
I = ∫(a/4 (4x3) + bx)/(x4 + c2) dx
= a/4 ∫(4x3)/(x4 + c2) dx + b∫x/((x2)2 + c2) dx
= a/4 ∫(4x3)/(x4 + c2) dx + b/2 ∫2x/((x2)2 + c2) dx
= a/4 log|x4 + c2| + b/2 I1 ……..(i)
Now,
I1 = ∫2x/((x2)2 + c2) dx
Put x2 = t
2xdx = dt
I1 = ∫1/((t)2 + c2) dx
= 1/c tan-1(t/c) + c1
l1 = 1/C tan-1(x2/c) + c1 …………(ii)
Now using equation (i) and (ii) we get,
Hence, I = a/4 log|x4 + c2| + b/2c tan(x2/c) + b
问题10.∫(x + 2)/(2x 2 + 6x + 5)dx
解决方案:
Given that I = ∫(x + 2)/(2x2 + 6x + 5) dx
Let us considered x + 2 = m d/dx (2x2 + 6x + 5) + n
= m(4x + 6) + n
x + 2 = (4m)x + (6m + n)
On comparing the coefficients of x,
So,
4m = 1
m = 1/4
6m + n = 2
6(1/4) + n = 2
n = 1/2
I = ∫(1/4(4x + 6)+1/2)/(2x2 + 6x + 5) dx)
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/2 ∫1/(2x2 + 6x + 5) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 3x + 5/2) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 2x(3/2) + (3/2)2 – (3/2)2 + 5/2) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + 1/4) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + (1/2)2) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 × (1/(1/2))tan-1((x + 3/2)/(1/2)) + c
As we know that ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c
Hence, I = 1/4 log|2x2 + 6x + 5| + 1/2 tan-1(2x + 3) + c
问题11∫((3sinx – 2)cosx)/(5 – COS 2x – 4sinx)DX
解决方案:
Given that I = ∫((3sinx – 2)cosx)/(5 – cos2x – 4sinx) dx
= ∫((3sinx – 2)cosx)/(5 – (1 – sin2x) – 4sinx) dx
= ∫((3sinx – 2)sinx)/(5 – 1 + sin2x – 4sinx) dx)
Now substitute sinx = t in the above equation
cosxdx = dt
So,
I = ∫(3t – 2)/(4 + t2 – 4t) dt
= ∫((3t – 2))/(t2 – 4t + 4) dt
= ∫(3t – 2)/(t – 2)2 dt
Now Integrate partial fractions.
(3t – 2)/((t – 2)2) = A/((t – 2)) + B/((t – 2)2)
= (A(t – 2) + B)/((t – 2)2)
= (At – 2A + B)/((t – 2)2)
3t – 2 = At – 2A + B
On comparing the coefficients, we have, A = 3
and -2A + B = -2
Now, on substituting the value of A = 3 in the above equation,
-2 × 3 + B = -2
-6 + B = -2
B = 6 – 2
B = 4
So, I = ∫(3t – 2)/(t – 2)2dt
= ∫(3t – 2)/(t – 2) dt + ∫4/(t – 2)2 dt
= 3log|t – 2| – 4(1/(t – 2)) + c
= 3log|t – 2| – 4(1/(t – 2)) + c
Now put the value of t = sinx, we have ,
I = 3log|sinx – 2| – 4(1/(sinx – 2)) + c
问题12.∫(5x – 2)/(1 + 2x + 3x 2 )dx
解决方案:
Given that I = ∫(5x – 2)/(1 + 2x + 3x2) dx
Let us considered 5x – 2 = A d/dx (1 + 2x + 3x2) + B
5x – 2 = A(2 + 6x) + B
5x – 2 = 6 × A + 2A + B
On comparing the Co-efficient we have, 6A = 5 and 2A + B = -2
A = 5/6
On substituting the value of A in 2A + B = -2, we n have,
2 * 5/6 + B = -2
10/6 + B = -2
B = -2 – 10/6
B = (-12 – 10)/6
B = (-22)/6
B = (-11)/3
5x – 2 = 5/6(2 + 6x) – 11/3
So, I = ∫(5x – 2)/(1 + 2x + 3x2) dx becomes,
I = ∫[5/6(2 + 6x) – 11/3]/(3x2 + 2x + 1) dx
= 5/6∫(2 + 6x)/(3x2+ 2x + 1) dx – 11/3∫dx/(3x2 + 2x + 1)
= 5/6 log(3x2 + 2x + 1) – 11/(3 × 3)∫dx/(x2 + 2/3x + 1/3) + c
= 5/6 log(3x2 + 2x + 1) – 11/9∫dx/(x2 + 2/3 x + (4/3)2 + 1/3 – (4/3)2) + c
= 5/6 log(3x2 + 2x + 1) – 11/9∫dx/((x + 1/3)2 +(√2/3)2) + c
= 5/6 log(3x2 + 2x + 1) – 11/9 × 1/(√2/3) tan-1(((x + 1/3)/(√2/3))] + C
Hence, I = 5/6 log(3x2 + 2x + 1) – 11/(3√2) tan-1[(3x + 1)/√2] + C