问题1.评估∫1/ a 2 -b 2 x 2 dx
解决方案：
Let us assume I = ∫ 1/ (a2-b2x2)dx
take 1/b2 common from above eqation
= 1/b2 ∫ 1/ (a2/b2-x2) dx
= 1/b2 ∫ 1/ (a/b)2-x2) dx
Integrate the above eq then, we get
= 1/b2 1/ 2(a/b) log|(a/b)+x/ (a/b)-x| + c [since ∫1/ a2-x2 dx = 1/2a log|x+a/x-a| + c]
= 1/2ab log|a+bx/ a-bx| + c
Hence, I = 1/2ab log |a+bx/ a-bx| + c
问题2.评估∫1 / a 2 x 2 -b 2 dx
解决方案：
Let us assume I = ∫ 1/ a2x2-b2 dx
take 1/a2 common from above equation
= 1/a2 ∫ 1/ x2-(b2/a2) dx
= 1/a2 ∫ 1/ x2-(b/a)2 dx
Integrate the above eq then, we get
= (1/a2) 1/(2b/a) log|x-(b/a)/x+(b/a)| + c [since ∫1/ x2-a2 dx = 1/2a log|x-a/x+a| + c]
= 1/2ab log|ax-b/ax+b| + c
Hence, I = 1/2ab log|ax-b/ax+b| + c
问题3.评估∫1 / a 2 x 2 + b 2 dx
解决方案：
Let us assume I = ∫ 1/ a2x2+b2 dx
take 1/a2 common from above equation
= 1/a2 ∫ 1/ x2+(b2/a2) dx
= 1/a2 ∫ 1/ x2+(b/a)2 dx
Integrate the above eq then, we get
= (1/a2) 1/(b/a)tan-1[x/(b/a)] + c [since ∫1/ x2+a2 dx = 1/a tan-1(x/a) + c]
= 1/ab tan-1(ax/b) + c
Hence, I = 1/ab tan-1(ax/b) + c
问题4.评估∫x 2 -1 / x 2 +4 dx
解决方案：
Let us assume I = ∫ x2-1/ x2+4 dx
We can write the above eq as below,
= ∫ x2-1+4-4/ x2+4 dx
= ∫ (x2+4)-4-1/ x2+4 dx
= ∫ (x2+4)-5/ x2+4 dx
= ∫ (x2+4)/ x2+4 dx – ∫ 5/ x2+4 dx
= ∫ dx – 5∫ 1/ x2+(2)2 dx
Integrate the above eq then, we get
= x – 5/2 tan-1(x/2) +c [since ∫1/ x2+a2 dx = 1/a tan-1(x/a) + c]
Hence I = x – 5/2 tan-1(x/2) +c
问题5.评估∫1 /√1+ 4x 2 dx
解决方案：
Let us assume I = ∫ 1/ √1+4x2 dx
= ∫ 1/ √1+(2x)2 dx (i)
Let 2x = t
2dx = dt
Put the above value in eq (i)
=1/2 ∫ 1/ √1+t2 dt
Integrate the above eq then, we get
=1/2 log|t+√t2+1| + c [since ∫1/ √x2+a2 dx = log|x+√x2+a2| +c]
=1/2 log|2x+√(2x)2+1| + c
Hence, I =1/2 log|2x+√4x2+1| + c
问题6.评估∫1/√A2 + B 2×2 DX
解决方案：
Let us assume I = ∫1/ √a2+b2x2 dx
= ∫1/ √a2+(bx)2 dx (i)
Let bx = t
bdx = dt
dx = dt/b
Put the above value in eq (i)
= 1/b ∫1/ √a2+(t)2 dt
Integrate the above eq then, we get
= 1/b log|t+√a2+t2|+ c [since ∫1/ √a2+x2 dx = log|x+√a2+x2| + c]
= 1/b log|bx+√a2+(bx)2|+ c
Hence, I = 1/b log|bx+√a2+b2x2|+ c
问题7.评估∫1/√A2 -B 2×2 DX
解决方案：
Let us assume I = ∫1/ √a2-b2x2 dx
= ∫1/ √a2-(bx)2 dx (i)
Let bx = t
bdx = dt
dx = dt/b
Put the above value in eq (i)
= 1/b ∫1/ √a2-(t)2 dt
Integrate the above eq then, we get
= 1/b sin-1(t/a)+ c [since ∫1/ √a2-x2 dx = sin-1 (x/a)+ c]
Hence, I = 1/b sin-1(bx/a)+ c
问题8.评估∫1/√(2-x) 2 +1 dx
解决方案：
Let us assume I = ∫1/ √(2-x)2+1 dx (i)
Let 2-x=t
-dx = dt
Put the above value in eq (i)
= -∫1/ √(t)2+(1)2 dt
Integrate the above eq then, we get
= – log |t+√(t)2+1| + c [since ∫1/ √x2+a2 dx = log|x+√x2+a2| +c]
= – log |(2-x)+√(2-x)2+1| + c
Hence, I = – log |(2-x)+√(2-x)2+1| + c
问题9.评估∫1/√(2-x) 2 -1 dx
解决方案：
Let us assume I = ∫1/ √(2-x)2-1 dx (i)
Let 2-x=t
-dx = dt
Put the above value in eq (i)
= -∫1/ √(t)2-(1)2 dt
Integrate the above eq then, we get
= – log |t+√(t)2-1| + c [since ∫1/ √x2-a2 dx = log|x+√x2-a2| +c]
= – log |(2-x)+√(2-x)2-1| + c
Hence, I = – log |(2-x)+√(2-x)2-1| + c
问题10.评估∫x 4 + 1 / x 2 +1 dx
解决方案：
Let us assume I = ∫ x4+1/ x2+1 dx
= ∫ (x2)2+(1)2/ x2+1 dx
= ∫ (x2+1)2-2x2/ x2+1 dx [a2 +b2 = (a+b)2-2ab]
= ∫ (x2+1)2/ x2+1 dx -∫ 2x2/ x2+1 dx
= ∫ (x2+1) dx – ∫ (2x2+2-2/ x2+1) dx
= ∫ (x2+1) dx – ∫ 2(x2+1)/ x2+1 dx +∫ 2/ x2+1 dx
= ∫ (x2+1) dx – ∫ 2 dx + 2∫ 1/ x2+1 dx
Integrate the above eq then, we get
= x3/3 + x – 2x + 2tan-1(x) + c [since ∫1/ x2+a2 dx = 1/a tan-1(x/a) + c]
= x3/3 – x + 2tan-1(x) + c
Hence, I = x3/3 – x + 2tan-1(x) + c