第 12 类 RD Sharma 解——第 19 章不定积分——练习 19.21
问题 1. ∫x/√(x²+6x+10) dx
解决方案:
Let l=∫x/√(x²+6x+10) dx
Let x=2 d/dx (x²+6x+10)+μ
=λ(2x+6)+μ
x=(2λ)x+6λ+μ
Comparing the coefficients of x,
2λ=1
λ=1/2
6λ+μ=0
6(1/2)+μ=0
μ=-3
so, l1=∫(1/2(2x+6)-3)/√(x²+6x+10) dx
=1/2 ∫ ((2x+6))/√(x²+6x+10) dx-3∣1/√(x²+2x(3)+(3)²-(3)²+10) dx
I1=1/2 ∫ (2x+6)/√(x²+6x+10) dx-3] 1/√((x+3)²+(1)²) dx
l1=1/2(2√(x²+6x+10))-3log|x+3+√((x+3)²+1)|+c
[∫1/√x dx=2√x+c, ∫1/√(x²+a²) dx=log|x+√(x²+a²)|+c]
I=√(x²+6x+10)-3log|x+3+√(x²+6x+10)|+c
问题 2. ∫ (2x+1)/√(x²+2x-1) dx
解决方案:
Let I= ∫ (2x+1)/√(x²+2x-1) dx
Let 2x +1=λ d/dx (x²+2x-1)+μ
=λ(2x+2)+μ
2x+1=(2λ)x+2λ+μ
Comparing the coefficients of x ,
2λ=2
λ=1
2λ+μ=1
2(1)+μ=1
μ=-1
so, I=∫((2x+2)-1)/√(x²+2x-1) dx
=∫((2x+2))/√(x²+2x-1) dx-∫1/√(x²+2x+(1)²-(1)²-1) dx
I=∫(2x+2)/√(x²+2x-1) dx-∫1/√((x+1)²-(√2)²) dx
I=(2√(x²+2x-1))-log|(x+1)+√((x+1)²-(√2)²)|+c
I=2√(x²+2x-1)-log|x+1+√(x²+2x-1)|+c
问题 3. ∫ (x+1)/√(4+5x-x²) dx
解决方案:
Let I=∫ (x+1)/√(4+5x-x²) dx
Let x+1=λ d/dx (4+5x-x²)+μ
=λ(5-2x)+μ
x=(-2λ)x+5λ+μ
Comparing the coefficients of x,
so,
-2λ=1
λ=-1/2
5λ+μ=1
5(-1/2)+μ=1
μ=7/2
I=∫ (-1/2(5-2x)+7/2)/√(4+5x-x²) dx
=-1/2 ∫ ((5-2x))/√(4+5x-x²) dx+7/2 ∫ 1/√(-[x²-5x-4]) dx
I=-1/2 ∫ (5-2x)/√(4+5x-x²) dx+7/2 ∫ 1/√(-[x²-2x(5/2)+(5/2)²-(5/2)²-4]) dx
I=-1/2 ∫ (5-2x)/√(4+5x-x²) dx+7/2 ∫ 1/√(-[(x-5/2)²-(√41/2)²]) dx
I=-1/2 ∫ (5-2x)/√(4+5x-x²) dx+7/2 ∫ 1/√((√41/2)²-(x-5/2)²]) dx
I=-1/2 (2√(4+5x-x²))+7/2 sin-1((x-5/2)/(√41/2))+c
I=-√(4+5x-x²)+7/2 sin-1((2x-5)/√41)+c
问题 4. ∫(6x-5)/√(3x²-5x+1) dx
解决方案:
Let I=∫(6x-5)/√(3x²-5x+1) dx
Let 3x²-5x+1=t
(6x-5)dx=dt
I=∫ dt/√t
=2√t+c
I=2√(3x²-5x+1)+c
问题 5. ∫(3x+1)/√(5-2x-x²) dx
解决方案:
Let I=∫(3x+1)/√(5-2x-x²) dx
Let 3x+1=λ d/dx (5-2x-x²)+μ
=λ(-2-2x)+μ
3x+1=(-2λ)×-2λ+μ
Comparing the coefficients of x,
-2λ=3
λ=-3/2
-2λ+μ=1
-2(-3/2)+μ=1
μ=-2
so, I=∫ (-3/2(-2-2x)-2)/√(5-2x-x²) dx
=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[x²+2x-5]) dx
I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[x²+2x+(1)²-(1)²-5]) dx
I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[(x+1)²-(√6)²]) dx
[∫ 1/√x dx=2√x+c,∫ 1/√(a²-x²) dx=sin-1(x/a)+c]
I=-3/2×2√(5-2x-x²)-2sin-1((x+1)/√6)+c
问题 6. ∫ x/√(8+xx²) dx
解决方案:
Let I =∫ x/√(8+x-x²) dx
Let x =λ d/dx (8+x-x²)+μ
=λ(1-2x)+μ
x=(-2λ)x+λ+μ
Comparing the coefficients of x,
so,
-2λ=1
λ=-1/2
λ+μ=0
(-1/2)+μ=0
μ=1/2
I=∫ (-1/2(1-2x)+1/2)/√(8+x-x²) dx
=-1/2 ∫ ((1-2x))/√(8+x-x²) dx+1/2 ∫ 1/√(-[x²-x-8]) dx
I=-1/2 ∫ (1-2x)/√(8+x-x²) dx+1/2 ∫ 1/(√(-[x²-2x(1/2)+(1/2)²-(1/2)²-8]) dx
I=-1/2 ∫ (1-2x)/√(8+x-x²) dx+1/2 ∫ 1/(√(-[(x-1/2)²-(33/4)²]) dx
I=-1/2 ∫ ((1-2x))/√(8+x-x²) dx+1/2 ∫ 1/√((√33/2)²-(x-1/2)²]) dx
I=-1/2×2√(8+x-x²)+1/2 sin-1((x-1/2)/(√33/2))+c
I=-√(8+x-x²)+1/2 sin-1((2x-1)/√33)+c
问题 7. ∫(x+2)/√(x²+2x-1) dx
解决方案:
Let l=∫(x+2)/√(x²+2x-1) dx
Let x+2=λ d/dx (x²+2x-1)+μ
x+2=λ(2x+2)+μ
x+2=(2λ)x+2λ+μ
Comparing the coefficients of x,
2λ =1 ∫
λ =1/2
2λ +μ=2
2*(1/2)+μ=2
μ=1
So, I = ∫(1/2(2x+2)+1)dx/√(x²+2x-1)
=1/2∫(2x+2)dx/√(x²+2x-1) + ∫1dx//√(x²+2x+1²-1²-1)
=1/2∫(2x+2)dx/√(x²+2x-1) + ∫1dx/(√(x+1)²-√2²)
I=1/2(2*(x²+2x-1)) +log|x+1+(√(x+)²-√2²)|+c
I=(x²+2x-1) +log|x+1+√(²+2x+1)|+c
问题 8. ∫(x+2)/√(x²-1) dx
解决方案:
Let x+2=A d/dx (x²-1)+B——————————-(i)
x+2=A(2x)+B
Equating the coefficients of x and constant term on both sides, we obtain
2A=1
A=1/2
B=2
From (1), we obtain
(x+2)=1/2(2x)+2
Then, ∫ (x+2)/√(x²-1) dx
=∫ (1/2(2x)+2)/√(x²-1) dx
=1/2 ∫ 2x/√(x²-1) dx+∫ 2/√(x²-1) dx———————–(ii)
In1/2 ∫ 2x/√(x²-1) dx,
let x²-1=t
2xdx=dt
1/2 ∫ 2x/√(x²-1) dx
=1/2 ∫ dt/√t
=1/2[2√t]
=√t
=√(x²-1)
Then, ∫2/√(x²-1) dx=2∫1/√(x²-1) dx=2log∣x+√(x²-1)
From equation (2), we obtain
∫(x+2)/√(x²-1) dx=√(x²-1)+2log|x+√(x²-1)|+c
问题 9. ∫ (x-1)/√(x²+1) dx
解决方案:
∫ (x-1)/√(x²+1) dx
=∫ (x-1)/√(x²+1) dx
=∫ x/√(x²+1)-1/√(x²+1) dx
=1/2 ∫ 2x/√(x²+1) dx-∫ 1/√(x²+1) dx [let t=x²+1, dt=2xdx]
=1/2 ∫ dt/√t-∫ 1/√(x²+1) dx
=1/2(2√t)-∫ 1/√(x²+1) dx
=√t-ln|x+√(x²+1)|+c
=√(x²+1)-ln|x+√(x²+1)|+c
问题 10. ∫ x/√(x²+x+1) dx
解决方案:
Let I =∫ x/√(x²+x+1) dx
Let x =λ d/dx (x²+x+1)+μ
=λ(2x+1)+μ
x=(2λ)×+λ+μ
Comparing the coefficients of x,
So,
=1/2 ∫ ((2x+1))/√(x²+x+1) dx-1/2 ∫ 1/√(x²+2x(1/2)+(1/2)²-(1/2)²+1) dx
I=1/2 ∫ (2x+1)/√(x²+x+1) dx-1/2 ∫1/(√((x+1/2)²-(√3/2)²)dx)
I=1/2×2√(x²+x+1)-1/2 log|x+1/2+√((x+1/2)²-(√3/2)²)|+c
I=√(x²+x+1)-1/2 log|x+1/2+√(x²+x+1)|+c
问题 11. ∫ (x+1)/√(x²+1) dx
解决方案:
Let I=∫ (x+1)/√(x²+1) dx
Let x+1=λ d/dx (x²+1)+μ
x+1=λ(2x)+μ
Comparing the coefficients of x,
2λ=1
λ=1/2
μ=1
I=∫ (1/2(2x)+1)/√(x²+1) dx
=1/2 ∫ (2x)/√(x²+1) dx+∫ 1/√(x²+1) dx
I=1/2×2√(x²+1)+log|x+√(x²+1)|+c
I=√(x²+1)+log|x+√(x²+1)|+c
问题 12. ∫ (2x+5)/√(x²+2x+5) dx
解决方案:
Let I= ∫ (2x+5)/√(x²+2x+5) dx
Let 2x +5=λ d/dx (x²+2x+5)+μ
=λ(2x+2)+μ
2x+5=(2λ)x+2λ+μ
Comparing the coefficients of x,
2λ=2
λ=1
2λ+μ=5
2(1)+μ=5
μ=3
so, l =∫ ((2x+2)+3)/√(x²+2x+5) dx
= ∫ (2x+3)/√(x²+2x+5) dx+3∫ 1/√(x²+2x+(1)²-(1)²+5) dx
I =∫ (2x+3)/√(x²+2x+5) dx+3∫ 1/√((x+1)²+(2)²) dx
I =2√(x²+2x+5)+3log|x+1+√((x+1)²+(2)²)|+c
I=2√(x²+2x+5)+3log|x+1+√(x²+2x+5)|+c
问题 13. ∫ (3x+1)/√(5-2x-x²) dx
解决方案:
Let I= ∫ (3x+1)/√(5-2x-x²) dx
Let 3x +1=λ d/dx (5-2x-x²)+μ
=λ(-2-2x)+μ
3x+1=(-2λ)x-2λ+μ
Comparing the coefficients of x ,
-2λ=3
λ=-3/2
-2λ+μ=1
-2(-3/2)+μ=1
μ=-2
Now, I=∫ (-3/2(-2-2x)-2)/√(5-2x-x²) dx
=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[x²+2x-5]) dx
I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[x²+2x+(1)²-(1)²+5]) dx
I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[(x+1)²-(√6)²]) dx
I=-3/2×2√(5-2x-x²)-2sin-1((x+1)/√6)+c
I=-3√(5-2x-x²)-2sin-1((x+1)/√6)+c
问题 14. ∫ √((1-x)/(1+x)) dx
解决方案:
Let I=∫ √((1-x)/(1+x)) dx
=∫ √((1-x)/(1+x)×(1-x)/(1-x)) dx
=∫ (1-x)/√(1-x²) dx
Let 1-x=λ d/dx (1-x²)+μ
=λ(-2x)+μ
1-x=(-2λ)x+μ
Comparing the coefficients of x,
-2λ=-1
λ=1/2
μ=1
I=∫ (1/2(-2x)+1)/√(1-x²) dx
=1/2 ∫ (-2x)/√(1-x²) dx+∫ 1/√(1-x²) dx
I=1/2×2√(1-x²)+sin-1x+c
I=√(1-x²)+sin-1x+c
问题 15. ∫ (2x+1)/√(x²+4x+3) dx
解决方案:
Let I= ∫ (2x+1)/√(x²+4x+3) dx
Let 2x +1=λ d/dx (x²+4x+3)+μ
=λ(2x+4)+μ
2x+1=(2λ)x+4λ+μ
Comparing the coefficients of x,
2λ=2
λ=1
4λ+μ=1
4*1+μ=1
μ=-3
I=∫((2x+4)-3)/√(x²+4x+3) dx
=∫(2x+4)/√(x²+4x+3) dx-3∫1/√(x²+2x(2)+(2)²-(2)²+3) dx
=∫(2x+4)/√(x²+4x+3) dx-3∫1/√((x+2)²-1) dx
I=2√(x²+4x+3)-3log|x+2+√((x+2)²-1)|+c
I=2√(x²+4x+3)-3log|x+2+√(x²+4x+3)|+c
问题 16. ∫ (2x+3)/√(x²+4x+5) dx
解决方案:
Let I=∫ (2x+3)/√(x²+4x+5) dx
Let 2x+3=λ d/dx (x²+4x+5)+μ
=λ(2x+4)+μ
2x+3=(2λ)x+4λ+μ
Comparing the coefficients of x,
2λ=2
λ=1
4λ+μ=3
4(1)+μ=3
μ=-1
Now, I=∫ ((2x+4)-1)/√(x²+4x+5) dx
= ∫ (2x+4)/√(x²+4x+5) dx-∫ 1/√(x²+2x(2)+(2)²-(2)²+5) dx
= ∫ (2x+4)/√(x²+4x+5) dx-∫ 1/√((x+2)²+(1)²) dx
I=2√(x²+4x+5)-log|x+2+√((x+2)²+1)|+c
I=2√(x²+4x+5)-log|x+2+√(x²+4x+5)|+c
问题 17. ∫(5x+3)/√(x²+4x+10) dx
解决方案:
∫(5x+3)/√(x²+4x+10) dx
let 5x+3=λ(2x+4)+μ
λ=5/2,
μ=-7
∫(λ(2x+4)+μ)/√(x²+4x+10) dx
=∫(5/2(2x+4)-7)/√(x²+4x+10) dx
=∫(5/2(2x+4))/√(x²+4x+10) dx-∫7/√(x²+4x+10) dx
=∫(5/2 dt)/√t-∫7/√((x+2)²+6) dx
=5√(x²+4x+10)-7log|(x+2)+√(x²+4x+10)|+c
问题 18. ∫(x+2)/√(x²+2x+3) dx
解决方案:
Let I=∫(x+2)/√(x²+2x+3) dx
x+2=A d/dx [x²+2x+3]+B
x+2=2Ax+2A+B
Comparing the co-efficients, we have, 2A=1 and 2A+B=2
A=1/2
Substituting the value of A in 2A+B=2, we have, 2×1/2+B=2
1+B=2
B=2-1
B=1
Thus we have, x+2=1/2[2x+2]+1
Hence, I=∫(x+2)/√(x²+2x+3) dx
=∫[1/2[2x+2]+1]/√(x²+2x+3) dx
=∫[1/2[2x+2]+1]/√(x²+2x+3) dx
=∫[1/2[2x+2]]/√(x²+2x+3) dx+∫dx/√(x²+2x+3)
=1/2∫([2x+2])/√(x²+2x+3) dx+∫dx/√(x²+2x+3)
Substituting t=x²+2x+3 and dt=2x+2 in the first integrand,
we have, I=1/2∫dt/√t+∫dx/√(x²+2x+3)
=1/2×2√t+∫dx/√(x²+2x+1+2)+c
=√t+∫dx/√((x+1)²+(√2)²)+c
I=√(x²+2x+3)+log[|x+1|+√((x+1)²+(√2)²)]+c
I=√(x²+2x+3)+log[|x+1|+√(x²+2x+3)]+c