问题21.∫dx/(x [6(logx)²+7logx+ 2])
解决方案:
Let =∫dx/(x[6(logx)²+7logx+2])
=∫1/(x(2logx+1)(3logx+2)) dx
Now,
Let 1/(x(2logx+1(3logx+2))=A/(x(2logx+1))+B/(x(3logx+2))
1=A(3logx+2)+B(2logx+1)
Put x=10-1/2
1=(1/2) A
A = 2n Put x = 10-2/3
1=(-1/3)B
B=-3
I=∫2xdx/(x(2logx+1))-∫3dx/(x(3logx+2))
=log|2logx+1|+log|3logx+2|+c
I=log|(2logx+1)/(3logx+2)|+c
问题22.∫1 / x(x n +1)dx
解决方案:
1/x(xn+1)
Multiplying numerator and denominator by xn-1, we obtain
1/x(xn+1) =xn-1/(xn-1 x(xn+1))=xn-1/(xn (xn+1))
Let xn=t
x n-1dx=dt
∫ 1/x(xn+1) dx=∫ xn-1/(xn (xn+1)) dx
=1/n ∫ 1/(t(t+1)) dt
Let 1/(t(t+1))=A/t+B/((t+1))
1=A(1+t)+Bt ——————————-(i)
Substituting t=0,-1 in equation (1), we obtain
A=1 and B=-1
∫ 1/(t(t+1)) =1/t -1/(1+t)
∫ 1/x(xn+1) dx =1/n ∫ {1/t-1/(t+1)}dx
=1/n[log|t|-log|t+1|]+c
=-1/n [log|xn |-log|xn+1|]+c [xn=t]
=1/n log|xn/(xn+1)|+Cc
问题23.∫x /(x²-a²)(x²-b²)dx
解决方案:
Let x/(x²-a²)(x²-b²) =(Ax+B)/(x²-a²) +(Cx+D)/(x²-b²)
x =(Ax+B)(x²-b²)+(Cx+D)(x²-a²)
x=(A+C)x3+(B+D)x²+(-Ab²-Ca²)x+(-Bb²-Da²)
A+C=0,
B+D=0,
-Ab²-Ca²=1,
-Bb²-Da²=0
We get B=0 , D=0 , C=1/(b²-a²) And A=-1/(b²-a²)
Thus,
I=-1/(b²-a²)∫x dx/(x²-a²) + 1/(b²-a²)∫x dx/(x²-b²)
I=-1/2(b²-a²) log|x²-a² |+1/2(b²-a²) log|x²-b² |+c
问题24.∫(x²+ 1)/(x²+ 4)(x²+ 25)dx
解决方案:
Consider the integral I=∫(x²+1)/(x²+4)(x²+25) dx
Let y=x²
Thus,
(x²+1)/(x²+4)(x²+25) =(y+1)/((y+4)(y+25))
(y+1)/((y+4)(y+25))=A/(y+4)+B/(y+25)
(y+1)/((y+4)(y+25)) = (A(y+25)+B(y+4))/((y+4)(y+25))
y+1=Ay+25A+By+4B
Comparing the coefficients, we have,
A+B=1 and 25A+4B=1
Solving the above equations, we have
A=(-1)/7 and B=8/7
Thus, ∫(x²+1)/(x²+4)(x²+25) dx
=∫((-1)/7)/(x²+4) dx+∫(8/7)/(x²+25) dx
=(-1)/7∫1/(x²+4) dx+8/7∫1/(x²+25) dx
=(-1)/7×1/2 tan-1x/2+8/7×1/5 tan-1x/5+C
=(-1)/14 tan-1x/2+8/35 tan-1x/5+C
问题25.∫(x 3 + x + 1)/(x²-1)dx
解决方案:
Let I =∫(x3+x+1)/(x2-1) dx
=∫(x+(2x+1)/(x²-1))dx
Now,
Let (2x+1)/(x²-1)=A/(x+1)+B/(x-1)
2x+1=A(x-1)+B(x+1)
Put x=1
3=2B
B=3/2
Put x=-1
-1=-2A
A=1/2
I=∫x dx+1/2∫dx/(x+1)+3/2∫dx/(x-1)
I=x²/2+1/2 log|x+1|+3/2 log|x-1|+c
问题26.∫(3x-2)/((x + 1)²(x + 3))dx
解决方案:
Let (3x-2)/((x+1)² (x+3))=A/(x+1)+B/((x+1)²)+C/((x+3))
3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)²
=(A+C)x²+(4A+B+2C)x+(3A+3B+C)
Equating similar terms, we get,
A+C=0
A=-C
4A+B+2C=3
B=-2C=3
3A+3B+C=-2
3B-2C=-2
Solving, we get,
B=-5/2 , C=-11/4 And A=11/4
Thus,
I=11/4 ∫ dx/(x+1)-5/2 ∫ dx/((x+1)²)-11/4 ∫ dx/(x+3)
I=11/4 log|x+1|+5/(2(x+1))-11/4 log|x+3|+c
问题27.∫(2x + 1)/((x + 2)(x-3)²)dx
解决方案:
Let (2x+1)/((x+2)(x-3)²)=A/(x+2)+B/(x-3)+C/((x-3)²)
2x+1=A(x-3)²+B(x+2)(x-3)+C(x+2)
=(A+B)x²+(-6A-B+C)x+(9A-6B+2C))
Equating similar terms, we get,
A+B=0
A=-B
-6A-B+C=2
5B+C=2
9A-6B+2C=1
-15B+2C=1
Solving, we get,
B=3/25 , C=7/5 And A=-3/25
Thus,
I=-3/25 ∫ dx/(x+2)+3/25 ∫ dx/(x-3)+7/5 ∫ dx/((x-3)²)
I=-3/25 log|x+2|+3/25 log|x-3|-7/(5(x-3))+c
问题28.∫(x²+ 1)/((x-2)²(x + 3))dx
解决方案:
Let (x²+1)/((x-2)² (x+3))=A/(x-2)+B/((x-2)²)+C/(x+3)
x²+1=A(x-2)(x+3)+B(x+3)+C(x-2)²
=(A+C)x²+(A+B-4C)x+(-6A+3B+4C)
Equating similar terms, we get,
A+C=1,
A+B-4C=0,
-6A+3B+4C=1
Solving, we get,
A=3/5,
B=1,
C=2/5
Thus,
I=3/5 ∫ dx/(x-2)+∫ dx/((x-2)²)+2/5 ∫ dx/(x+3)
I=3/5 log|x-2|-1/((x-2))+2/5 log|x+3|+c
问题29.∫x /(((x-1)²(x + 2))dx
解决方案:
Let x/((x-1)² (x+2))=A/(x-1)+B/((x-1)²)+C/((x+2))
x=A(x-1)(x+2)+B(x+2)+C(x-1)²
Substituting x=1, we obtain
B=1/3
Equating the coefficients of x² and constant term, we obtain
A+C=0
-2A+2B+C=0
On solving, we obtain
A=2/9 and C=(-2)/9
x/((x-1)² (x+2)) = 2/(9(x-1))+1/(3(x-1)²)-2/(9(x+2))
∫x/((x-1)² (x+2)) dx =2/9∫1/((x-1)) dx+1/3∫1/((x-1)²) dx-2/9∫1/((x+2)) dx
=2/9 log|x-1|+1/3 ((-1)/(x-1))-2/9 log|x+2|+c
=2/9 log|(x-1)/(x+2)|-1/(3(x-1))+c
问题30.∫x²/(((x-1)(x + 1)²)dx
解决方案:
Let x²/((x-1)(x+1)²)=A/(x-1)+B/(x+1)+C/((x+1)²)
x²=A(x+1)²+B(x-1)(x+1)+C(x-1)
=(A+B)x²+(2A+C)x+(A-B-C)
Equating similar terms,
A+B=1,
2A+C=0,
A-B-C=0
Solving, we get,
A=1/4,
B=3/4,
C=-1/2
Thus,
I =1/4 ∫ dx/(x-1)+3/4 ∫ dx/(x+1)-1/2 ∫ dx/((x+1)²)
=1/4 log|x-1|+3/4 log|x+1|+1/2(x+1)+c
I =1/4 log|x-1|+3/4 log|x+1|+1/2(x+1)+c
问题31.∫(x²+ x-1)/(((x + 1)²(x + 2))dx
解决方案:
Let (x²+x-1)/((x+1)² (x+2))=A/((x+1))+B/((x+1)²)+C/((x+2))
x²+x-1=A(x+1)(x+2)+B(x+2)+C(x+1)²
=(A+C)x²+(3A+B+2C)x+(2A+2B+C))
Equating similar terms
A+C=1 , 3A+B+2C=1 And 2A+2B+C=-1
Solving, we get, A=0 , B=-1 And C=1
Thus,
I =0*∫ dx/(x+1)+(-1)∫ dx/((x+1)²)+1*∫ dx/((x+2))
=+1/(x+1)+log|x+2|+c
I =1/(x+1)+log|x+2|+c
问题32.∫(2x²+ 7x-3)/(x²(2x + 1))dx
解决方案:
Let (2x²+7x-3)/(x² (2x+1))=A/x+B/x² +C/(2x+1)
2x²+7x-3=Ax(2x+1)+B(2x+1)+Cx²
Equating similar terms, we get,
2A+C=2 , A+2B=7 And B=-3
Solving, we get, A=13 And C=-24
Thus,
I= ∫ 13dx/x-∫3dx/x² -24∫dx/(2x+1)
I=13log|x|+3/x-12log|2x+1|+c
问题33.∫(5x²+ 20x + 6)/(x 3 +2x²+ x)dx
解决方案:
Let I=∫(5x²+20x+6)/(x3+2x²+x) dx
=∫(5x²+20x+6)/(x(x+1)²) dx
Now,
Let
(5x²+20x+6)/(x(x+1)²)=A/x+B/(x+1)+C/((x+1)²)
5x²+20x+6=A(x+1)²+Bx(x+1)+Cx-1
Equating similar terms, we get,
A+B=5 , 2A+B+C=20 And A=6
Solving, we get, B=-1 And C=9
Thus,
I=∫6dx/x-1∫dx/(x+1)+9∫dx/((x+1)²)
I=6log|x|-log|x+1|-9/(x+1)+c
问题34.∫18 /(((x + 2)(x²+ 4))dx
解决方案:
Let 18/((x+2)(x²+4))=A/(x+2)+(Bx+C)/(x²+4)
18=A(x²+4)+(8x+c)(x+2)
18=(A+8)x²+(2B+C)x+(4A+2C)
Equating similar terms, we get,
A+B=0 , 2B+C=0 And 4A+2C=18
Solving, we get,
A=9/4, B=-9/4 And C=9/2
Thus,
I=9/4 ∫ dx/(x+2)+(-9/4) ∫ x/(x²+4) dx+9/2 ∫ dx/(x²+4) [∫ dx/(x²+a²)=1/a tan-1x/a]
I=9/4 log|x+2|-9/8 log|x²+4|+9/4 tan-1(x/2)+c [∵∫▒ dx/(x^2+a^2)=1/a tan^(-1)x/9])
问题35.∫5 /(((x²+ 1)(x + 2))dx
解决方案:
Let 5/((x²+1)(x+2))=(Ax+B)/(x²+1)+C/(x+2)
5=(Ax+B)(x+2)+C(x²+1)
Equating similar terms, we get,
A+C=0 , 2A+B=0 And 2B+C=5
Solving, we get,
A=-1 , B=2 And C=1
Thus,
I =∫ (-x+2)/(x²+1) dx+∫ dx/(x+2)
=∫ (-xdx)/(x²+1)+2∫ dx/(x²+1)+∫ dx/(x+2)
=-1/2 log|x²+1|+2tan-1x+log|x+2|+c
问题36.∫x/(((x + 1)(x²+ 1))dx
解决方案:
Let x/((x+1)(x²+1))=A/(x+1)+(Bx+C)/(x²+1)
x=A(x²+1)+(Bx+C)(x+1)
Equating similar terms, we get,
A+B=0 , B+C=1 And A+C=0
Solving, we get, A=-1/2 , B=1/2 And C=1/2
Thus,
I=-1/2∫dx/(x+1)+1/2∫xdx/(x²+1)+1/2∫dx/(x²+1)
I=-1/2 log|x+1|+1/4 log|x²+1|+1/2 tan-1x+c
问题37.∫dx/(1 + x +x²+ x 3 )
解决方案:
Let I=∫dx/(1+x+x²+x3)
I=∫dx/((x²+1)(x+1))
Now,
Let 1/((x²+1)(x+1))=(Ax+B)/(x²+1)+C/(x+1)
1=(Ax+B)(x+1)+C(x²+1)
Equating similar terms, we get,
A+C=0, A+B=0 And B+C=1
Solving, we get,
A=-1/2, B=1/2 And C=1/2
Thus,
I=-1/2∫xdx/(x²+1)+1/2∫dx/(x²+1)+1/2∫dx/(x+1)
I=-1/4 log|x²+1|+1/2 tan-1x+1/2 log|x+1|+c
问题38.∫1 /(((x + 1)²(x²+ 1))dx
解决方案:
Let 1/((x+1)²(x²+1))=A/(x+1)+B/((x+1)²)+(Cx+D)/(x²+1)
1 =A(x+1)(x²+1)+B(x²+1)+(Cx+D)(x+1)²
=(A+C)x3+(A+B+2C+D)x²+(A+C+2D)x+(A+B+D)
Equating similar terms, we get,
A+C=0, A+B+2C+D=0, A+C+2D=0 And A+B+D=1
Solving, we get,
A=1/2, B=1/2, C=-1/2 And D=0
Thus,
I=1/2 ∫ dx/(x+1)+1/2 ∫ dx/((x+1)²)-1/2 ∫ xdx/(x²+1)
I=1/2 log|x+1|-1/(2(x+1))-1/4 log|x²+1|+c
问题39.∫2x/(x 3 -1)dx
解决方案:
Let I=∫2x/(x3-1) dx
=∫2x/((x-1)(x²+x+1)) dx
Now,
Let 2x/((x-1)(x²+x+1))=A/((x-1))+(Bx+C)/(x²+x+1)
2x=A(x²+x+1)+(Bx+C)(x-1)
=(A+B)x²+(A-B+C)x+(A-C)
Equating similar terms,
A+B=0, A-B+C =2, And A-C=0
Solving, we get,
A=2/3, B=-2/3, C=2/3
Thus,
I=2/3 ∫ dx/(x-1)-2/3 ∫ ((x-1)dx)/(x²+x+1)
=2/3 ∫ dx/(x-1)-2/3*1/2 ∫ ((2x-2)dx)/(x²+x+1)
I=2/3 ∫ dx/(x-1)-1/3 ∫ (2x+1)/(x²+x+1) dx+∫ dx/(x²+x+1))
I =2/3 ∫ dx/(x-1)-1/3 ∫ (2x+1)/(x²+x+1) dx+∫ dx/(x²+x+1)
=2/3 ∫ dx/(x-1)-1/3 ∫ (2x+1)/(x²+x+1) dx+∫ dx/((x+1/2)²+(√3/2)²)
=2/3 log|x-1|-1/3 log|x²+x+1|+2/√3 tan-1(x+1/2)/(√3/2)+c
Hence,
I=2/3 log|x-1|-1/3 log|x²+x+1|+2/√3 tan-1((2x+1)/√3)+c
问题40.∫1 /(x²+ 1)(x²+ 4)dx
解决方案:
Let 1/(x²+1)(x²+4) =(Ax+B)/((x²+1))+(Cx+D)/(x²+4)
1= (Ax+B)(x²+4)+(Cx+D)(x²+1)
=(A+C)x3+(B+D)x²+(4A+C)x+4B+D
Equating similar terms, we get
A+C=0, B+D=0, 4A+C=0 And 4B+D=1
Solving, we get, A=0, B=1/3, C=0, D=-1/3
Thus,
I=∫ (1/3 dx)/((x²+1))-∫ (1/3 dx)/((x²+4))
=1/3 tan-1x-1/6 tan-1(x/2)+c [∫ dx/(x²+a²)=1/a tan-1x/a]
I=1/3 tan-1x-1/6 tan-1(x/2)+c