第12类RD Sharma解–第19章不定积分–练习19.31

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问题1.∫(x ² +1)/(x ⁴ + x ² +1)dx

解决方案：

We have,

∫(x² + 1)/(x⁴ + x² + 1)dx

= ∫x²(1 + 1/x²)/x²(x² + 1 + 1/x²)dx

= ∫(1 + 1/x²)/(x² + 1 + 1/x²)dx

= ∫(1 + 1/x²)/(x² + 1/x² − 2 + 2 + 1)dx

= ∫(1 + 1/x²)/[(x − 1/x)² + 3]dx

Let x − 1/x = t.

Differentiating both sides, we get,

(1 + 1/x²)dx = dt

So, our equation becomes,

= ∫1/(t² + 3)dt

= (1/√3) tan^-1(t/√3) + c

= (1/√3) tan^-1[(x² − 1)/√3x] + c

问题2.∫√(cotθ)dθ

解决方案：

We have,

∫√(cotθ)dθ

Let cotθ = x²

Differentiating both sides, we get,

=> −cosec²θdθ = 2xdx 

=> dθ = −2xdx/cosec²θ

=> dθ = −2xdx/(1 + cot²θ)

=> dθ = −2xdx/(1 + x⁴)

So, our equation becomes,

= −∫2x²/(1 + x⁴)]dx

= −∫2/(x² + 1/x²)dx

= −∫(1 + 1/x² + 1 − 1/x²)/(x² + 1/x² + 2 − 2)dx

= −∫(1 + 1/x²)/[(x − 1/x)² + 2]dx − ∫(1 − 1/x²)/[(x + 1/x)² − 2]dx

Let x − 1/x = t, so we get, (1 + 1/x²)dx = dt

Let x + 1/x = y, so we get, (1 − 1/x²)dx = dy

So, our equation becomes, 

= −∫dt/(t² + 2) − ∫dy/(y² − 2)

= −(1/√2) tan^-1[t/√2] − (1/2√2) log |(y − √2)/(y + √2)| + c

= −(1/√2) tan^-1[(x² − 1)/√2x] − (1/2√2) log |(x² + 1 − √2x)/(x² + 1 + √2x)| + c

= −(1/√2) tan^-1[(cotθ − 1)/√(2cotθ)] − (1/2√2) log |(cotθ + 1 − √(2cotθ))/(cotθ + 1 + √(2cotθ))| + c

问题3.∫(x ² + 9)/(x ⁴ + 81)dx

解决方案：

We have, 

∫(x² + 9)/(x⁴ + 81)dx

= ∫x²(1 + 9/x²)/x²(x² + 81/x²)dx

= ∫(1 + 9/x²)/[(x − 9/x)² + 18]dx 

Let x−9/x = t, so we have, (1 + 9/x²)dx = dt 

So, our equation becomes,

= ∫dt/(t² + 18)

= (1/3√2) tan^-1(t/3√2) +c

= (1/3√2) tan^-1[(x² − 9)/3√2x] + c

问题4.∫1/(x ⁴ + x ² +1)dx

解决方案：

We have,

∫1/(x⁴ + x² + 1)dx

= ∫(1/x²)/(x² + 1 + 1/x²)dx

= (1/2)∫(1 + 1/x² − 1 + 1/x²)/(x² + 1 + 1/x²)dx

= (1/2)∫(1 + 1/x²)/[(x − 1/x)² + 3]dx − (1/2)∫(1 − 1/x²)/(x + 1/x)²− 1)dx

Let x − 1/x = t, so we get, (1 + 1/x²)dx = dt and 

Let x + 1/x = y, so we get, (1 − 1/x²)dx = dy.

So, our equation becomes,

= (1/2)∫dt/(t² + 3) − (1/2)∫dy/(y² − 1)

= (1/2√3) tan^-1(t/√3) − (1/4) log |(y − 1)/(y + 1)| + c

= (1/2√3) tan^-1[(x² − 1)/√3x] − (1/4) log |(x² + 1 − x)/(x² + 1 + x)| + c

问题5.∫(x ² − 3x +1)/(x ⁴ + x ² +1)dx

解决方案：

We have,

∫(x² − 3x + 1)/(x⁴ + x² + 1)dx

= ∫(1 − 3/x + 1/x²)/(x² + 1 + 1/x²)dx

= ∫(1 + 1/x²)/[(x − 1/x)² + 3]dx − ∫3x/(x⁴ + x² + 1)dx

Let x − 1/x = t, so we get, (1 + 1/x²)dx = dt

Let x² = y, so we get, 2xdx = dy 

So, our equation becomes,

= ∫dt/(t² + 3) − (3/2)∫dy/(y² + y + 1)

= ∫dt/(t² + 3) − (3/2)∫dy/[(y+1/2)² + 3/4]

= (1/√3) tan^-1(t/√3) − (3/2)(2/√3)tan^-1[(y + 1/2)/(√3/2)] + c

= (1/√3) tan^-1(t/√3) − √3tan^-1[(2y + 1)/√3] + c

= (1/√3) tan^-1[(x² − 1)/√3] − √3tan^-1[(2x² + 1)/√3] + c

问题6.∫(x ² +1)/(x ⁴ − x ² +1)dx

解决方案：

We have,

∫(x² + 1)/(x⁴ − x² + 1)dx

= ∫(1 + 1/x²)/(x² − 1 + 1/x²)dx

= ∫(1 + 1/x²)/[(x − 1/x)² + 1]dx

Let x − 1/x = t, so we get, (1 + 1/x²)dx = dt

So, the equation becomes,

= ∫dt/(t² + 1)

= tan⁻¹t + c

= tan⁻¹[(x² − 1)/x] + c

问题7.∫(x ² − 1)/(x ⁴ +1)dx

解决方案：

We have,

∫(x² − 1)/(x⁴ + 1)dx

= ∫(1 − 1/x²)/(x² + 1/x²)dx

= ∫(1 − 1/x²)/[(x + 1/x)² − 2]dx

Let x + 1/x = t, so we get, (1 − 1/x²)dx = dt

So, the equation becomes,

= ∫dt/(t² − 2)

= (1/2√2) log |(t − √2)/(t + √2)| + c

= (1/2√2) log |(x² + 1 − √2x)/(x² + 1 + √2x)| + c

问题8.∫(x ² +1)/(x ⁴ + 7x ² +1)dx

解决方案：

We have,

∫(x² + 1)/(x⁴ + 7x² + 1)dx

= ∫(1 + 1/x²)/(x² + 7 + 1/x²)dx

= ∫(1 + 1/x²)/[(x − 1/x)² + 9]dx

Let x − 1/x = t, so we get, (1 + 1/x²)dx = dt

So, the equation becomes,

= ∫dt/(t² + 9)

= (1/3)tan⁻¹(t/3) + c

= (1/3)tan⁻¹[(x² − 1)/3x] + c

问题9.∫(x − 1) ² /(x ⁴ + x ² +1)dx

解决方案：

We have,

∫(x − 1)²/(x⁴ + x² + 1)dx

= ∫(x² − 2x + 1)/(x⁴ + x² + 1)dx

= ∫(1 − 2/x + 1/x²)/(x² + 1 + 1/x²)dx

= ∫(1 + 1/x²)/[(x − 1/x)² + 3]dx − ∫2x/(x⁴ + x² + 1)dx

Let x − 1/x = t, so we get, (1 + 1/x²)dx = dt

Let x² = y, we get, 2xdx = dy

So, our equation becomes,

= ∫dt/(t² + 3) − ∫dy/(y² + y + 1)

= ∫dt/(t² + 3) − ∫dy/[(y + 1/2)² + 3/4]

= (1/√3) tan⁻¹[(x² − 1)/√3x] − (2/√3) tan⁻¹[(2x² + 1)/√3] + c

问题10.∫1/(x ⁴ + 3x ² +1)dx

解决方案：

We have,

∫1/(x⁴ + 3x² + 1)dx

= ∫(1/x²)/(x² + 3 + 1/x²)dx

= (1/2)∫(1 + 1/x² − 1 + 1/x²)/(x² + 3 + 1/x²)dx

= (1/2)∫(1 + 1/x²)/[(x − 1/x)² + 5]dx − (1/2)∫(1 − 1/x²)/[(x + 1/x)² + 1]dx

Let x − 1/x = t, we get, (1 + 1/x²)dx = dt

Let x + 1/x = y, we get, (1 − 1/x²)dx = dy

So, the equation becomes,

= (1/2)∫dt/(t² + 5) − (1/2)∫dy/(y² + 1)

= (1/2√5) tan⁻¹(t/√5) − (1/2) tan⁻¹y + c

= (1/2√5) tan⁻¹[(x² − 1)/√5x] − (1/2) tan⁻¹[(x² + 1)/x] + c

问题11.∫1/(罪⁴ x +罪² x cos ² x + cos ⁴ x)dx

解决方案：

We have, 

∫1/(sin⁴x + sin²x cos²x + cos⁴x)dx

= ∫(1/cos⁴x)/[(sin⁴x + sin²x cos²x + cos⁴x)/(cos⁴x)]dx

= ∫(sec⁴x)/(tan⁴x + tan²x + 1)dx

= ∫[(1 + tan²x)sec²x]/(tan⁴x + tan²x + 1)dx

Let tan x = t, so we get sec²xdx = dt

So, the equation becomes,

= ∫(1 + t²)/(t⁴ + t² + 1)dt

= ∫(1 + 1/t²)/(t² + 1/t² + 1)dt

= ∫(1 + 1/t²)/[(t − 1/t)² + 3]dt

Let t − 1/t = y, so we get, (1 + 1/t²)dt = dy

So, the equation becomes,

= ∫dy/(y² + 3)

= (1/√3) tan⁻¹(y/√3) + c

= (1/√3) tan⁻¹[(t−1/t)/√3] + c

= (1/√3) tan⁻¹[(tan x − 1/tan x)/√3] + c

= (1/√3) tan⁻¹[(tan x − cot x)/√3] + c