问题41.∫x²/(x²+ 1)(3x²+ 4)dx
解决方案:
Let x²/(x²+1)(3x²+4) =(Ax+B)/(x²+1) +(Cx+D)/(3x²+4)
x²=(Ax+8)(3x²+4)+(Cx+D)(x²+1)
=(3A+C)x3+(3B+D)x²+(4A+C)x+4B+D
Equating similar terms, we get,
3A+C=0,
3B+D=1,
4A+C=0,
4B+D=0
Solving, we get,
A=0,
B=-1,
C=0,
D=4
Thus,
I =∫ (-dx)/(x²+1) +∫ 4dx/((3x²+4))
=-tan-1x+4/3 ∫ dx/(x²+(2/√3)²)
=-tan-1x+(4/3)*(√3/2)tan-1((√3 x)/2)+c
I =2/√3 tan-1((√3 x)/2)-tan-1x+c
问题42.∫(3x + 5)/(x 3 -x²-x+ 1)dx
解决方案:
∫ (3x+5)/(x3-x²-x+1) dx
Let (3x+5)/((x-1)² (x+1))=A/(x-1)+B/((x-1)²)+C/(x+1)
3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)²
Put x=1
B=4
Put x=-1
C=1/2
Put x=0
A=-1/2
Therefore
∫ (3x+5)/((x-1)² (x+1)) dx
=-1/2 ∫ dx/(x-1)+4∫ dx/((x-1)²)+1/2 ∫ dx/(x+1)
=-1/2 ln|(x-1)|-4/((x-1))+1/2 ln|(x+1)|+c
=1/2 ln|(x+1)/(x-1)|-4/((x-1))+c
问题43.∫(x 3 -1)/(x 3 + x)dx
解决方案:
Let I =∫ (x3-1)/(x3+x) dx
=∫(1-(x+1)/(x3+x))dx
=∫ dx-∫ (x+1)/(x3+x) dx
Let (x+1)/x(x²+1) =A/x+(Bx+C)/(x²+1)
x+1 =A(x²+1)+(Bx+C)x
=(A+8)x²+(B+C)x+A
Equating similar terms, we get,
A+B=0,
C=1,
A=1
Solving, we get,
A=1,
B=-1,
C=1
Thus,
I=-∫ dx/x-∫ (-x+1)/(x²+1) dx+∫ dx
I =x-log|x|+1/2 log|x²+1|-tan-1x+c
I=x-log|x|+1/2 log|x²+1|-tan-1x+c
问题44.∫(x²+ x + 1)/((x + 1)²(x + 2))dx
解决方案:
∫ (x²+x+1)/((x+1)² (x+2)) dx
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
Putting x=-1
B=1
Putting x=-2
C=3
Putting x=0
A=-2,
Therefore,
∫ (x²+x+1)/((x+1)² (x+2)) dx
=-2∫ dx/(x+1)+∫ dx/((x+1)²)+3∫ dx/(x+2)
=-2ln|x+1|-1/(x+1)+3ln|x+2|+c
问题45.∫1 / x(x 4 +1)dx
解决方案:
Let 1/x(x4+1) =A/x+(Bx3+Cx2+Dx+E)/(x4+1)
1=A(x4+1)+(Bx3+Cx²+Dx+E)x
=(A+B)x4+Cx3+Dx²+Ex+A
Equating similar terms, we get,
A+B=0,
C=0,
D=0,
E=0,
A=1,
B=-1
Thus,
I =∫ dx/x+∫ -(x3 dx)/(x4+1)
=log|x|-1/4 log|x4+1|+c
I =1/4 log|x4/(x4+1)|+c
问题46.∫1/(x(x 3 +8))dx
解决方案:
I=∫1/(x(x3+8)) dx
Arranging the above equation,
I=∫ x²/(x3 (x3+8)) dx
=1/3∫(3x²)/(x3 (x3+8)) dx
Now substituting x3=t, we have, 3x3 dx=dt
I=1/3∫dt/(t(t+8))
Integrand by partial fractions. Thus,
1/(t(t+8))=A/t+B/(t+8)
1/(t(t+8))=(A(t+8)+Bt)/(t(t+8))
1=A(t+8)+Bt
1=At+8A+Bt
Comparing the coefficients, we have,
A+B=0,
8A=-1/8
Therefore,
I=1/3 ∫ dt/(t(t+8))
=1/3∫ {(1/8)/t – (1/8)/(t+8)} dt
=(1/3)*(1/8) ∫ dt/t – (1/3)*(1/8)∫ dt/(t+8)
=(1/24)* log | t |-(1/24)* log| t+8 |+c
=(1/24) log| x3 |-(1/24) log| x3+8|+c
=(3/24) log|x3|- (1/24)log|x3+8|+c
I=(1/8) log| x3| – (1/24)log|x3+8|+c
问题47.∫3 /(((1-x)(1 +x²))dx
解决方案:
Let 3/((1-x)(1+x²))=A/(1-x)+(Bx+C)/(1+x²)
3 =A(1+x²)+(Bx+C)(1-x)
=(A-B)x²+(B-C)x+(A+C)
Equating similar terms, we get,
A-B=0,
B-C=0,
A+C=3
Solving we get,
A=C=3/2 and B=3/2
Thus,
I=3/2 ∫ dx/(1-x)+3/2 ∫ xdx/(1+x²)+3/2 ∫ dx/(1+x²)
=-3/2 log|1-x|+3/2 log|1+x² |+3/2 tan-1x+c
I=3/4 [log|(1+x²)/((1-x)²)|+2tan-1x+c
问题48.∫(cosx)/(((1-sinx) 3 (2 +sinx))dx
解决方案:
Let sinx=t
cosx=dt
∫ (cosx)/((1-sinx)3(2+sinx))=∫ 1/((1-t)3 (2+t)) dt
Let f(t)=1/((1-t)3 (2+t))
Then, suppose
1/((1-t)3 (2+t)) = A/(1-t)+B/((1-t)²)+c/((1-t)3)+D/((2+t))
1=A(1-t)² (2+t)+B(1-t)(2+t)+C(2+t)+D(1-t)3
Put t=1
1=3C
C=1/3
Put t=-2
1=27D
D=1/27
Similarly, we can find that A=(-1)/27 and B=(+1)/9
∫ 1/((1-t)3 (2+t)) dt
=(-1)/27 ∫ 1/(1-t) dt+1/9 ∫ dt/((1-t)²)+1/3 ∫ dt/((1-t)3)+1/27 ∫ dt/(2+t)
=(-1)/27 log|1-t|+1/(9(1-t))+1/(6(1-t)²)+1/27 log|2+t|+c
Putting t=sinx , we get
∫ (cosx)/((1-sinx)3 (2+sinx)) dx
=(-1)/27 log|1-sinx|+1/(9(1-sinx))+1/(6(1-sinx)²)+1/27 log|2+sinx|+c
问题49.∫(2x²+ 1)/(x²(x²+ 4))dx
解决方案:
I=∫(2x²+1)/(x² (x²+4)) dx
Now let us separate the fraction (2x²+1)/(x² (x²+4)) through partial fractions.
Substitute x²=t, then (2x²+1)/(x² (x²+4))=(2t+1)/(t(t+4))
(2t+1)/(t(t+4))=A/t+B/(t+4)
(2t+1)/(t(t+4))=(A(t+4)+Bt)/(t(t+4))
2t+1=A(t+4)+Bt
2t+1=At+4A+Bt
Comparing the coefficients, we have, A+B=2 and 4A=1
A=1/4 and B=7/4
(2x²+1)/(x² (x²+4))=1/(4x²)+7/4(x²+4)
Thus ,we have
I= ∫ (2x²+1)/(x²(x²+4)) dx
=1/4 ∫ dx/x² dx +7/4 ∫ dx/(x²+4) dx
=-1/4x +(7/4)*(1/2) tan-1(x/2)+c
I=-1/4x+(7/8)tan-1(x/2)+c
问题50.∫cosx /(1-sinx)/(2-sinx)dx
解决方案:
∫ cosx/(1-sinx)/(2-sinx) dx
Let 1-sinx=t and
-cos x dx = dt
Therefore
-∫ dt/(t(1+t))=-∫ (1/t-1/(t+1))dt
=ln|(t+1)|-ln|t|+c
=ln|(t+1)/t|+c
=ln|(2-sinx)/(1-sinx)|+c
问题51.∫(2x + 1)/((x-2)(x-3))dx
解决方案:
Let (2x+1)/((x-2)(x-3))=A/((x-2))+B/(x-3)
2x +1=A(x-3)+B(x-2)
=(A+B)x+(-3A-2B)
Equating similar terms, we get,
A+B=2, and -3A-2B=1
Thus,
I=-5∫ dx/(x-2)+7∫ dx/(x-3)
=-5log|x-2|+7log|x-3|+c
I=log|((x-3)7/((x-2)5)|+c
问题52.∫1 /(((x²+ 1)(x²+ 2))dx
解决方案:
Let x²=y
Then 1/((y+1)(y+2))=A/(y+1)+B/(y+2)
1 =A(y+2)+B(y+1)
=(A+B)y+(2A+B)
Equating similar terms, we get,
A+B=0, and 2A+B=1
Solving, we get,
Thus,
I=∫ dx/(x²+1)-∫ dx/(x²+2)
I=tan-1x-1/√2 tan-1x/√2+c
问题53.∫1 / x(x 4 -1)dx
解决方案:
∫ 1/x(x4-1) dx
Let 1/x(x4-1) =A/x+B/(x+1)+C/(x-1)+D/(x²+1)
1=A(x+1)(x-1)(x²+1)+Bx(x-1)(x²+1)+Cx(x+1)(x²+1)+Dx(x+1)(x-1))
Put x=0
A=-1,
Put x=1
C=1/4
Put x=-1
B=1/4
Put x=2
D=1/4
Therefore
∫ 1/x(x4-1) dx
=-∫ 1/x dx+1/4 ∫ dx/(x+1)+1/4 ∫ dx/(x-1)+1/4 ∫ dx/(x²+1)
=-ln|x|+1/4 ln|(x+1)|+1/4 ln|(x-1)|+1/4 ln|(x²+1)|+c
=1/4 ln|(x4-1)/x4 |+c
问题54.∫1 /(x 4 -1)dx
解决方案:
∫ 1/(x4-1) dx
Let 1/(x4-1) =A/(x+1)+B/(x-1)+C/(x²+1)
1=A(x-1)(x²+1)+B(x+1)(x²+1)+C(x+1)(x-1)
Put x=1
B=1/
Put x=-1
A=-1/4
Put x=0
C=-1/2
Therefore,
∫ 1/(x4-1) dx
=-1/4 ∫ dx/(x+1)+1/4 ∫ dx/(x-1)-1/2 ∫ dx/(x²+1)
=-1/4 ln|(x+1)|+1/4 ln|(x-1)|-1/2 tan-1x+c
=1/4 ln|(x-1)/(x+1)|-1/2 tan-1x+c
问题55.∫dx /(cosx(5-4sinx))dx
解决方案:
Let I=∫ dx/(cosx(5-4sinx))
=∫ (cosxdx)/(cos²x(5-4sinx))
=∫ (cosxdx)/((1-sin²x)(5-4sinx))
Let sinx=t
cos x dx = dt
I=∫ dt/((1-t²)(5-4t))
Now,
Let 1/((1-t²)(5-4t))=A/(1-t)+B/(1+t)+C/(5-4t)
1=A(1+t)(5-4t)+B(1-t)(5-4t)+C(1-t²)
Put t=1
1=2A
A=1/2
Put t=-1
1=18B
B=1/18
Put t=5/4
1=-9C/16
C=-16/9
Thus,
I=1/2∫dt/(1-t)+1/18∫dt/(1+t)-16/9∫dt/(5-4t)
=-1/2 log|1-t|+1/18 log|1+t|+4/9 log|5-4t|+c
Hence,
I=-1/2 log|1-sinx|+1/18 log|1+sinx|+4/9 log|5-4sinx|+c
问题56.∫1 /(sinx(3 +2cosx))dx
解决方案:
Let I =∫ 1/(sinx(3+2cosx)) dx
=∫ (sinxdx)/(sin²x(3+2cosx))
=∫ (sinxdx)/((1-cos²x)(3+2cosx))
Let cosx=t
-sinxdx=dt
I=∫ dt/((t²-1)(3+2t))
Now,
Let 1/((t²-1)(3+2t))=A/(t-1)+B/(t+1)+C/(3+2t)
1=A(t+1)(3+2t)+B(t-1)(3+2t)+C(t²-1)
Put t=1
1=10A
A=1/10
Put t=-1
1=-2B
B=-1/2
Put t=-3/2
1=5/4C
C=4/5
Thus,
I=1/10∫dt/(t-1)-1/2∫dt/(t+1)+5/4∫dt/(3+2t)
=1/10 log|t-1|-1/2 log|t+1|+2/5 log|3+2t|+c
Hence,
I=1/10 log|cosx-1|-1/2 log|cosx+1|+2/5 log|3+2cosx|+c
问题57.∫1/(sinx+sin2x)dx
解决方案:
Let I=∫1/(sinx+sin2x) dx
=∫ dx/(sinx+2sinxcosx)
=∫ (sinxdx)/((1-cos²x)+2(1-cos²x)cosx)
Let cosx=t
-sinxdx=dt
I =∫ dt/((t²-1)+2(t²-1)t)
=∫ dt/((t²-1)(1+2t))
Let ∫1/((t²-1)(1+2t))=A/(t-1)+B/(t+1)+C/(1+2t)
1=A(t+1)(1+2t)+B(t-1)(1+2t)+c(t²-1)
Put t=1
1=6A
A=1/6
Put t=-1
1=2B
B=1/2
Put t=-1/2
1=-3/4 C
C=-4/3
Thus,
I=1/6∫dt/(t-1)+1/2∫dt/(t+1)-4/3∫dt/(1+2t)
=1/6 log|t-1|+1/2 log|t+1|-2/3 log|1+2t|+c
Hence,
I=1/6 log|cosx-1| +1/2 log|cosx+1| -2/3 log|1+2cosx| +c
问题58.∫(x + 1)/ x(1 + xe x )dx
解决方案:
Let I=∫(x+1)/x(1+xex) dx
=∫((x+1)(1+xex-xex))/x(1+xex) dx
=∫((x+1)(1+xex))/x(1+xex) dx – ∫((x+1)(xex))/x(1+xex) dx
=∫((x+1))/x dx-∫(ex (x+1))/(1+xex) dx
=∫((x+1)ex)/(xex) dx-∫(ex (x+1))/(1+xex) dx
=log|xex |-log|1+xex |+c
I=log|(xex)/(1+xex)|+c
问题59.∫(x²+ 1)(x²+ 2)/(x²+ 3)(x²+ 4)dx
解决方案:
f(x)=(x²+1)(x²+2)/(x²+3)(x²+4)
Now,
((x²+1)(x²+2)/(x²+3)(x²+4)
=(x4+3x2+2)/(x4+7x2+12)
=((x4+7x²+12)-4x²-10)/(x4+7x²+12)
=1-(4x²+10)/(x4+7x²+12)
Now,
(4x²+10)/(x4+7x²+12)
=(4x²+10)/(x²+3)(x²+4)
Let (4x²+10)/(x²+3)(x²+4) =(Ax+B)/(x²+3)+(Cx+D)/(x²+4)
4x²+10=(Ax+B)(x²+4)+(Cx+D)(x²+3)
Let x=0, we get
10=48+3D—————————–(i)
If x=1, we get
14=5(A+B)+4(C+D)=5A+5B+4C+4D——————(ii)
if x=-1, we get
14=5(-A+B)+4(-C+D)=-5A+5B-4C+4D—————-(iii)
Applying (ii) and (iii) we get,
28=10B+8D
1=5B+4D ——————————–(iv)
From (i) we get,
10=4B+3D
Multiplying equation (iv) by 3 and (i) by 4 and subtracting, we get
42-40=15B-16B
2=-B
B=-2
Putting value of B in (i), we get
10=4(-2)+3D
(10+8)/3=D
D=6
Comparing coefficients of x3 in
4x²+10=(Ax+B)(x²+4)+(Cx+4)(x²+3),
we get
0=A+C
Comparing coefficients of x, we get
0=4A+3C
A=C=0
f(x)=1-(-2)/(x²+3)-6/(x²+4)
=1+2/(x²+3)-6/(x²+4)
∫ f(x)dx=∫ 1+2/(x²+3)-6/(x²+4) dx
=x+2/√3 tan-1x/√3-3tan-1x/2+c
问题60.∫(4×4 + 3)/((x²+ 2)(x²+ 3)(x²+ 4))dx
解决方案:
let x²=y
(4x4+3)/(x²+2)(x²+3)(x²+4)
=(4y²+3)/((y+2)(y+3)(y+4))
Now,
Let (4y²+3)/((y+2)(y+3)(y+4))=A/(y+2)+B/(y+3)+C/(y+4)
4y²+3 =A(y+3)(y+4)+B(y+2)(y+4)+c(y+2)(y+3)
=(A+B+C)y²+(7A+6A+5C)y+12A+8B+6C
Equating similar terms,
A+B+C=4,
7A+6A+5C=0,
12A+8B+6C=3
Solving, we get
A=19/2,
B=-39,
C=67/2
Thus,
I=19/2 ∫ dx/(x²+2)+(-39)∫ dx/(x²+3)+67/2 ∫ dx/(x²+4)
I=19/(2√2) tan-1(x/√2)-39/√3 tan-1(x/√3)+67/4 tan-1(x/2)+c
Hence,
I=19/(2√2) tan-1(x/√2)-39/√3 tan-1(x/√3)+67/4 tan-1(x/2)+c
问题61.∫x 4 /(((x-1)(x²+ 1))dx
解决方案:
x4/((x-1)(x²+1))=x4/(x3-x²+x-1)
=(x(x3-x²+x-1)+1(x3-x²+x-1)+1)/((x3-x²+x-1))
=x+1+1/((x-1)(x²+1))
Now,
1/((x-1)(x²+1))=A/(x-1)+(Bx+C)/(x²+1)
1=A(x²+1)+(8x+C)(x-1)
Put x=1
1=2A
A=1/2
Put x=0
1=A-C
C=A-1=-1/2
Put x=-1
1=2A+2B-2C
=2(A-C)+2B
1=2+2B
2B=-1
B=-1/2
∫ x4/((x-1)(x²+1)) dx
=∫ xdx+∫ 1dx+1/2 ∫ 1/(x-1) dx-1/2 ∫ (x+1)/(x²+1) dx
=x²/2+x+1/2 log|x-1|-1/4 log|(x²+1)|-1/2 tan-1x+c