问题1.评估∫秒2 x / 1 –棕褐色2 x dx
解决方案：
Let us assume I = ∫ sec2x/ 1 – tan2x dx …..(i)
Now, put tan x = t
sec2x dx = dt
So, put all these values in eq(i)
= ∫ dt/ 12 – t2
On integrating the above equation then, we get
= 1/ 2(1) log|1 + t/1 – t| + c
Since, ∫ 1/ a2 – x2 dx = 1/ 2a log|a + x/a – x| + c]
Hence, I = 1/2 log|1 + tanx/1 – tanx| + c
问题2.评估∫e x / 1 + e 2x dx
解决方案：
Let us assume I = ∫ ex/ 1 + e2x dx …..(i)
Now, put ex = t
ex dx = dt
So, put all these values in eq(i)
= dt/ 1 + t2
On integrating the above equation then, we get
= tan-1t + c
Since, ∫ 1/ 1 + x2dx = tan-1x + c
Hence, I = tan-1ex + c
问题3.评估∫cosx / sin 2 x + 4sinx + 5 dx
解决方案：
Let us assume I = ∫ cosx/ sin2x + 4sinx + 5 dx …..(i)
Now, put sinx = t
cosx dx = dt
So, put all these values in eq(i)
= ∫ dt/ t2 + 4t + 5
= ∫ dt/ t2 + 2t(2) + (2)2 – (2)2 + 5
= ∫ dt/ (t + 2)2 + 1 …..(ii)
Again, Put t + 2 = u
dt = du
Now, put all these values in eq(ii)
= ∫ du/ u2 + 1
On integrating the above equation then, we get
= tan-1u + c
Since, ∫1/ x2 + 1 dx = tan-1x + c
= tan-1(t + 2) + c
Hence, I = tan-1(sinx + 2) + c
问题4.评估∫e x / e 2x + 5e x + 6 dx
解决方案：
Let us assume I = ∫ ex/e2x + 5ex + 6 dx …..(i)
Now, put ex = t
ex dx = dt
So, put all these values in eq(i)
= ∫ dt/ t2 + 5t + 6
= ∫ dt/ t + 2t(5/2) + (5/2)2 – (5/2)2 + 6
= ∫ dt/ (t + 5/2)2 – 1/4 …..(ii)
Put t + 5/2 = u
dt = du
Now, put the above value in eq(ii)
= ∫ du/ u2 – (1/2)2
On integrating the above equation then, we get
= 2/2 log|u – (1/2)/u + (1/2)| + c
Since, ∫ 1/ x2 – a2 dx = 1/ 2alog|x – a/x + a| + c
= log|2u – 1/2u + 1| + c
= log|2(t + 5/2) – 1/2(t + 5/2) + 1| + c
Hence, I = log|ex + 2/ex + 3| + c
问题5.评估∫e 3x / 4e 6x – 9 dx
解决方案：
Let us assume I = ∫ e3x/ 4e6x– 9 dx …..(i)
Now, put e3x = t
3e3x dx = dt
e3x dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ 4t2 – 9
= 1/12 ∫ dt/ t2 – (3/2)2
On integrating the above equation then, we get
= 1/12 x 1/ 2(3/2) log|t – 3/2/t + 3/2| + c
Since, ∫1/ x2 – a2 dx = 1/2a log|x – a/x + a| + c]
= 1/36 log|2t – 3/2t + 3| + c
Hence, I = 1/36 log|2e3x – 3/2e3x + 3| + c
问题6.评估∫dx / e x + e -x
解决方案：
Let us assume I = ∫ dx/ex + e-x
= ∫ dx/ex + 1/ex
= ∫ exdx/ (ex)2 + 1 …..(i)
Now, put ex = t
exdx = dt
Now, put the above value in eq(i)
= ∫ dt/ t2 + 1
On integrating the above equation then, we get
= tan-1t + c
Since ∫ 1/ 1 + x2 dx = tan-1x + c
Hence, I = tan-1(ex) + c
问题7.评估∫x / x 4 + 2x 2 + 3 dx
解决方案：
Let us assume I = ∫ x/ x4 + 2x2 + 3 dx …..(i)
Now, put x2 = t
2x dx = dt
x dx = dt/2
Now, put the above value in eq(i)
= 1/2 ∫ dt/ t2 + 2t + 3
= 1/2 ∫ dt/ t2 + 2t + 1 – 1 + 3
= 1/2 ∫ dt/ (t + 1)2 + 2 …..(ii)
Now put t + 1 = u
dt = du
So, put the above value in eq(ii)
= 1/2 ∫ du/ u2 + (√2)2
On integrating the above equation then, we get
= 1/2 x 1/√2 tan-1(u/√2) + c
Since ∫1/ x2 + a2dx = 1/a tan-1(x/a) + c
= 1/2√2 tan-1(t + 1/ √2) + c
Hence, I = 1/2√2 tan-1(x2 + 1/ √2) + c
问题8.评估∫3×5/1 +×12 DX
解决方案：
Let us assume I = ∫ 3x5/ 1 + x12 dx
= ∫ 3x5/ 1 + (x6)2dx …..(i)
Now, put x6 = t
6x5dx = dt
x5dx = dt/6
Now, put the above value in eq(i)
= 3/6 ∫ dt/ 1 + t2
On integrating the above equation then, we get
= 1/2 tan-1(t) + c
Since ∫ 1/ x2 + 1 dx = tan-1x + c
Hence, I = 1/2 tan-1(x6) + c
问题9.评估∫x 2 / x 6 – a 6 dx
解决方案：
Let us assume I = ∫ x2/ x6 – a6 dx
= ∫ x2/ (x3)2 – (a3)2 dx …..(i)
Now, put x3 = t
3x2 dx = dt
x2 dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ t2 – (a3)2
On integrating the above equation then, we get
= 1/3 x 1/2a3 log|t – a3/t + a3| + c
Since ∫1/ x2 – a2 dx = 1/2a log|x – a/x + a| + c
= 1/6a3 log|x3 – a3/x3 + a3| + c
Hence, I = 1/6a3 log|x3 – a3/x3 + a3| + c
问题10.评估∫x 2 / x 6 + a 6 dx
解决方案：
Let us assume I = ∫ x2/ x6 + a6 dx
= ∫ x2/ (x3)2 + (a3)2 dx …..(i)
Now, put x3 = t
3x2 dx = dt
x2 dx = dt/3
Now, put the above value in eq(i)
= 1/3 ∫ dt/ t2 + (a3)2
On integrating the above equation then, we get
= 1/3 x (1/a3) tan-1(t/a3) + c
Since, ∫1/ x2 + a2 dx = 1/a tan-1(x/a) + c
Hence, I = 1/3a3 tan-1(x3/a3) + c
问题11.评估∫1 / x(x 6 +1)dx
解决方案：
Let us assume I = ∫ 1/ x(x6 + 1) dx
= ∫ x5/ x6(x6 + 1) dx …..(i)
Now, put x6 = t
6x5 dx = dt
x5 dx = dt/6
Now, put the above value in eq(i)
= 1/6 ∫dt/ t(t + 1)
= 1/6 ∫dt/ t2 + t
= 1/6 ∫dt/ t2 + 2t(1/2) + (1/2)2 – (1/2)2
= 1/6 ∫dt/ (t + 1/2)2 – (1/2)2 …..(ii)
Let t + 1/2 = u
dt = du
So, put the above value in eq(ii)
= 1/6 ∫du/ (u)2 – (1/2)2
On integrating the above equation then, we get
= 1/6 x 1/ 2(1/2) log|u – (1/2)/u + (1/2)| + c
Since ∫ 1/ x2 – a2dx = 1/2a log|x – a/x + a| + c
= 1/6 log|{(t + 1/2) – 1/2}/(t + 1/2) + 1/2| + c
Hence, I = 1/6 log|x6/ x6 + 1| + c
问题12.评估∫x /(x 4 – x 2 +1)dx
解决方案：
Let us assume I = ∫ x/ (x4 – x2 + 1) dx …..(i)
Let x2 = t
2x dx = dt
x dx = dt/2
Now, put the above value in eq(i)
= 1/2 ∫dt/ t2 – t + 1
= 1/2 ∫dt/ t2 – 2t(1/2) + (1/2)2 – (1/2)2 + 1
= 1/2 ∫dt/ (t – 1/2)2 + (3/4) …..(ii)
Let t – 1/2 = u
dt = du
So, put the above value in eq(ii)
= 1/2 ∫du/ (u)2 + (√3/2)2
On integrating the above equation then, we get
= 1/2 x 1/(√3/2) tan-1(u/(√3/2)) + c
Since, ∫ 1/ x2 + a2dx = 1/a tan-1(x/a) + c
= 1/√3 tan-1(t – 1/2/ (√3/2)) + c
Hence, I = 1/√3 tan-1(2x2 – 1/ √3) + c
问题13.评估∫x /(3x 4 – 18x 2 + 11)dx
解决方案：
Let us assume I = ∫ x/ (3x4 – 18x2 + 11) dx
= 1/3 ∫ x/ (x4 – 6x2 + 11/3) dx …..(i)
Let x2 = t
2x dx = dt
x dx = dt/2
So, put the above value in eq(i)
= 1/3 x 1/2 ∫dt/ t2 – 6t + 11/3
= 1/6 ∫dt/ t2 – 2t(3) + (3)2 – (3)2 + 11/3
= 1/6 ∫dt/ (t – 3)2 – (16/3) …..(ii)
Let t – 3 = u
dt = du
Now, put the above value in eq(ii)
= 1/6 ∫du/ (u)2 – (4/√3)2
On integrating the above equation then, we get
= 1/6 x 1/ 2(4/√3) log|u – (4/√3)/u + (4/√3)| + c
Since, ∫ 1/ x2 – a2dx = 1/2a log|x – a/x + a| + c
= √3/48 log|(t – 3 – 4/√3)/(t – 3 + 4/√3)| + c
Hence, I = √3/48 log|(x2 – 3 – 4/√3)/(x2 – 3 + 4/√3)| + c
问题14.评估∫e x /(1 + e x )(2 + e x )dx
解决方案：
Let us assume I = ∫ ex/ (1 + ex)(2 + ex) dx …..(i)
Let ex = t
ex dx = dt
So, put the above value in eq(i)
= ∫ dt/ (1 + t)(2 + t)
= ∫dt/ (1 + t) – ∫dt/(2 + t)
On integrating the above equation then, we get
= log|1 + t| – log|2 + t| + c
= log|1 + t/2 + t| + c
Hence, I = log|1 + ex/2 + ex| + c