问题1.∫1/ [(x − 1)√(x + 2)] dx
解决方案:
We have,
∫1/[(x − 1)√(x + 2)]dx
Let x + 2 = t2, so we get, xdx = 2tdt
So, the equation becomes,
= ∫2t/(t2 − 3)(t)dt
= 2∫dt/(t2 − 3)
= (2/2√3) log |(t − √3)/(t + √3)| + c
= (1/√3) log |(√(x − 2) − √3)/(√(x − 2) + √3)| + c
问题2。∫1/ [(x − 1)√(2x + 3)] dx
解决方案:
We have,
∫1/[(x − 1)√(2x + 3)]dx
Let 2x + 3 = t2, so we have, 2dx = 2tdt,
=> dx = tdt
So, the equation becomes,
= ∫t/[(t2 − 3 − 2)/2](t)dt
= 2∫dt/(t2 − 5)
= (2/2√5) log |(t − √5)/(t + √5)| + c
= (1/√5) log |(√(2x + 3) − √5)/(√(2x + 3) + √5)| + c
问题3.∫(x + 1)/ [(x − 1)√(x + 2)] dx
解决方案:
We have,
∫(x + 1)/[(x − 1)√(x + 2)]dx
= ∫(x − 1 + 2)/[(x − 1)√(x + 2)]dx
= ∫(x − 1)/[(x − 1)√(x + 2)]dx + ∫2/[(x − 1)√(x + 2)]dx
= ∫(dx/√(x + 2)] + 2∫dx/[(x − 1)√(x + 2)]
In second part, let x + 2 = t2, so we get, xdx = 2tdt
So, the equation becomes,
= ∫(dx/√(x + 2)] + ∫2t/(t2 − 3)(t)dt
= ∫(dx/√(x + 2)] + 2∫dt/(t2 − 3)
= 2√(x + 2) + c1 + (4/2√3) log |(t − √3)/(t + √3)| + c2
= 2√(x + 2) + (2/√3) log |(√(x − 2) − √3)/(√(x − 2)+√3)| + c
问题4∫x2 / [(X – 1)√(X + 2)] DX
解决方案:
We have,
∫x2/[(x − 1)√(x + 2)]dx
= ∫(x2 − 1 + 1)/[(x − 1)√(x + 2)]dx
= ∫(x − 1)(x + 1)/[(x − 1)√(x + 2)]dx + ∫dx/[(x − 1)√(x + 2)]
= ∫(x + 1)/[√(x + 2)]dx + ∫dx/[(x − 1)√(x + 2)]
= ∫[(x + 2) − 1]/[√(x + 2)]dx + ∫dx/[(x − 1)√(x + 2)]
= ∫√(x + 2)dx − ∫dx/[√(x + 2)] + ∫dx/[(x − 1)√(x + 2)]
In third part, let x + 2 = t2, so we get, xdx = 2tdt
So, the equation becomes,
= ∫√(x + 2)dx − ∫dx/[√(x + 2)] + ∫2t/(t2 − 3)(t)dt
= ∫√(x + 2)dx − ∫dx/[√(x + 2)] + 2∫dt/(t2 − 3)
= (2/3)(x + 2)3/2 + c1 − 2√(x + 2) + c2 + (2/2√3) log |(t − √3)/(t + √3)| + c3
= (2/3)(x + 2)3/2 − 2√(x + 2) + (1/√3) log |(√(x − 2) − √3)/(√(x − 2) + √3)| + c
问题5.∫x/ [(x − 3)√(x + 1)] dx
解决方案:
We have,
∫x/[(x − 3)√(x + 1)]dx
= ∫[(x − 3) + 3]/[(x − 3)√(x + 1)]dx
= ∫dx/[√(x + 1)] + 3∫dx/[(x − 3)√(x + 1)]
For second part, let x + 1 = t2, so we get, dx = 2tdt.
So, the equation becomes,
= ∫dx/[√(x + 1)] + 3∫2tdt/[(t2 − 4)(t)]
= 2√(x + 1) + c1 + (3/2) log |(t − 2)/(t + 2)| + c2
= 2√(x + 1) + (3/2) log |(√(x + 1) − 2)/(√(x + 1) + 2)| + c
问题6.∫1/ [(x 2 +1)√x] dx
解决方案:
We have,
∫1/[(x2 + 1)√x]dx
Let x = t2, so we have, dx = 2tdt
So, the equation becomes,
= 2∫t/[(t4 + 1)(t)]dt
= 2∫dt/(t4 + 1)
= 2∫(t/t2)/(t2 + 1/t2)dt
= ∫[1 + 1/t2 − (1 − 1/t2)]/(t2 + 1/t2)dt
= ∫(1 + 1/t2)/[(t − 1/t)2 + 2]dt − ∫(1 − 1/t2)/[(t + 1/t)2 − 2]dt
Let t − 1/t = y, so we have, (1 + 1/t2)dt = dy
Let t + 1/t = z, so we have, (1 − 1/t2)dt = dz
So, the equation becomes,
= ∫dy/(y2 +2) − ∫dz/(z2 − 2)
= (1/√2) tan−1(y/√2) − (1/2√2) log |(z − √2)/(z + √2)| + c
= (1/√2) tan−1[(t2−1)/√2t] − (1/2√2) log |[x + 1 − √(2x)]/[x + 1 + √(2x)]| + c
= (1/√2) tan−1[(x−1)/√(2x)] − (1/2√2) log |[x + 1 − √(2x)]/[x + 1 + √(2x)]| + c
问题7.∫x/ [(x 2 + 2x + 2)√(x + 1)] dx
解决方案:
We have,
∫x/[(x2 + 2x + 2)√(x + 1)]dx
Let x + 1 = t2, so we have, dx = 2tdt
So, the equation becomes,
= 2∫(t2 − 1)(t)/[(t4 + 1)(t)]dt
= 2∫(t2 − 1)/(t4 + 1)dt
= 2∫(1 − 1/t2)/[(t + 1/t)2 − 2]dt
Let t + 1/t = y, so we have, (1 − 1/t2)dt = dy
So, the equation becomes,
= 2∫dy/(y2 − 2)
= (2/2√2) log |(y − √2)/(y + √2)| + c
= (1/√2) log |(t2 + 1 − √2t)/(t2 + 1 + √2t)| + c
= (1/√2) log |[x + 2 − √(2x + 2)]/[x + 2 + √(2x + 2)]| + c
问题8.∫1/ [(x − 1)√(x 2 +1)] dx
解决方案:
We have,
∫1/[(x − 1)√(x2 + 1)]dx
Let x − 1 = 1/t, so we have dx = (−1/t2)dt
So, the equation becomes,
= −∫(1/t2)/[(1/t)√[(1 + 1/t)2 + 1]]dt
= −∫dt/√(2t2 + 2t + 1)
= −(1/√2)∫dt/√(t2 + t + 1/2)
= −(1/√2)∫dt/√[(t + 1/2)2 + 1/4]
= −(1/√2) log |(t + 1/2) + √[(t + 1/2)2 + 1/4]| + c
= −(1/√2) log |(1/(x − 1) + 1/2) + √[(1/(x−1) + 1/2)2 + 1/4]| + c
问题9.∫1/ [(x + 1)√(x 2 + x + 1)] dx
解决方案:
We have,
∫1/[(x + 1)√(x2 + x + 1)]dx
Let x + 1 = 1/t, so we have dx = (−1/t2)dt
So, the equation becomes,
= −∫(1/t2)/[(1/t)√(1/t2 + 1/t − 1)]dt
= −∫dt/√(1 + t − t2)
= −∫dt/√[5/4 − (1/4 − t + t2)]
= −∫dt/√[5/4 − (t − 1/2)2]
= − sin−1[(t − 1/2)/(√5/2)] + c
= − sin−1[(2t − 1)/√5] + c
= − sin−1[(1 − x)/[√5(x + 1)]] + c
问题10.∫1/ [(x 2 − 1)√(x 2 +1)] dx
解决方案:
We have,
∫1/[(x2 − 1)√(x2 + 1)]dx
Let x = 1/t, so we get, dx = (−1/t2)dt
So, the equation becomes,
= −∫(1/t2)/[(1/t2 − 1)√(1/t2 + 1)]dt
= −∫t/[(1 − t2)√(1 + t2)]dt
Let 1 + t2 = y2, so we have, 2tdt = 2ydy
=> tdt = ydy
So, the equation becomes,
= ∫ydy/(y2 − 2)y
= ∫dy/(y2 − 2)
= (1/2√2) log |(y − √2)/(y + √2)| + c
= (1/2√2) log |(y − √2)/(y + √2)| + c
= (1/2√2) log |[√(1 + t2) − √2]/[√(1 + t2) + √2]| + c
= −(1/2√2) log |[√2x + √(x2 + 1)]/[√2x − √(x2 + 1)]| + c
问题11.∫x/ [(x 2 + 4)√(x 2 +1)] dx
解决方案:
We have,
∫x/[(x2 + 4)√(x2 + 1)]dx
Let x2 + 1 = t2, so we get, 2xdx = 2tdt
=> xdx = tdt
So, the equation becomes,
= ∫t/(t2 + 3)(t)dt
= ∫dt/(t2 + 3)
= (1/√3) tan−1(t/√3) + c
= (1/√3) tan−1[√(x2 + 1)/√3] + c
问题12.∫1/ [(1 + x 2 )√(1 − x 2 )] dx
解决方案:
We have,
∫1/[(1 + x2)√(1 − x2)]dx
Let x = 1/t, so we get, dx = (−1/t2)dt
So, the equation becomes,
= −∫(1/t2)/[(1/t2 + 1)√(1 − 1/t2)]dt
= −∫t/[(t2 + 1)√(t2 − 1)]dt
Let t2 − 1 = y2, so we get, 2tdt = 2ydy
=> tdt = ydy
So, the equation becomes,
= −∫y/[(y2 + 2)(y)]dy
= −∫1/(y2 + 2)dy
= −(1/√2) tan−1(y/√2) + c
= −(1/√2) tan−1(√(t2 − 1)/√2) + c
= −(1/√2) tan−1(√(1 − x2)/√2x) + c
问题13.∫1/ [(2x 2 + 3)√(x 2 − 4)] dx
解决方案:
We have,
∫1/[(2x2 + 3)√(x2 − 4)]dx
Let x = 1/t, so we have dx = (−1/t2)dt
So, the equation becomes,
= −∫(1/t2)/[(2/t2 + 3)√(1/t2 − 4)]dt
= −∫t/[(2 + 3t2)√(1−4t2)]dt
Let 1 − 4t2 = y2, so we get, −8tdt = 2ydy
So, the equation becomes,
= (1/4) ∫y/[(11 − 3y2)y/4]dy
= (1/3) ∫1/(11/3 − y2)dy
= (1/2√33) log |[y − √(11/3)]/[y + √(11/3)]| + c
= (1/2√33) log |[√(1 − 4t2) − √(11/3)]/[√(1 + 4t2) + √(11/3)]| + c
= (1/2√33) log |[√(11x) + √(3x2 − 12)]/[√(11x) − √(3x2 − 12)]| + c
问题14.∫x/ [(x 2 + 4)√(x 2 + 9)] dx
解决方案:
We have,
∫x/[(x2 + 4)√(x2 + 9)]dx
Let x2 + 9 = y2, so we have 2xdx = 2ydy
=> xdx = ydy
So, the equation becomes,
= ∫y/[(y2 − 5)y]dy
= ∫1/(y2 − 5)dy
= (1/2√5) log |(y − √5)/(y + √5)| + c
= (1/2√5) log |(√(x2 + 9) − √5)/(√(x2 + 9) + √5)| + c