重谢RMB,几道简单的求不定积分的题目!

太严重了吧，你来到香港再说。你的字体挺工整，相信你念书蛮勤力。
1、∫ [2^(x + 1) - 5^(x - 1)]/10^x dx
= ∫ 2^(x + 1)/10^x dx - ∫ 5^(x - 1)/10^x dx
= 2∫ 2^x/10^x dx - (1/5)∫ 5^x/10^x dx
= 2∫ (1/5)^x dx - (1/5)∫ (1/2)^x dx
= 2 · (1/5)^x/ln(1/5) - 1/5 · (1/2)^x/ln(1/2) + C
= (- 2/ln5)(1/5)^x - (1/5)(- 1/ln2)(1/2)^x + C
= (1/(5ln2))(1/2)^x - (2/ln5)(1/5)^x + C
公式：∫ a^x dx = a^x/lna + C

2、∫ cos²(x/2) dx
= (1/2)∫ 2cos²(x/2) dx
= (1/2)∫ (1 + cosx) dx
= (x + sinx)/2 + C
公式：cos2x = 2cos²x - 1

3、∫ dx/(x² - 2x + 5)
= ∫ dx/[(x - 1)² + 4]
= ∫ d(x - 1)/[(x - 1)² + 2²]
= (1/2)arctan[(x - 1)/2] + C
公式：∫ dx/(x² + a²) = (1/a)arctan(x/a) + C

4、∫ cos3x · sin2x dx
= (1/2)∫ [sin(3x + 2x) - sin(3x - 2x)] dx
= (1/2)∫ (sin5x - sinx) dx
= (1/2)(- 1/5 · cos5x + cosx) + C
= (1/10)(5cosx - cos5x) + C
公式：cosAsinB = (1/2)[sin(A + B) - sin(A - B)]

5、∫ sin²(x/2) dx
= (1/2)∫ 2sin²(x/2) dx
= (1/2)∫ (1 - cosx) dx
= (x - sinx)/2 + C
公式：cos2x = 1 - 2sin²x

6、∫ dx/(sin²xcos²x)
= ∫ dx/(1/2 · 2sinxcosx)²
= ∫ dx/[(1/2)²(sin2x)²]
= 4∫ csc²2x dx
= 2∫ csc²2x d(2x)
= 2 · (- cot2x) + C
= - 2cot2x + C
公式：sin2x = 2sinxcosx

7、∫ dx/(1 + cos2x)
= ∫ dx/[1 + (2cos²x - 1)]
= (1/2)∫ sec²x dx
= (1/2)tanx + C
公式：cos2x = 2cos²x - 1

8、∫ dx/(1 - cos2x)
= ∫ dx/[1 - (1 - 2sin²x)] dx
= (1/2)∫ csc²x dx
= (1/2)(- cotx) + C
= - (1/2)cotx + C
公式：cos2x = 1 - 2sin²x

1、令x = sinθ，dx = cosθ dθ
∫ [3 - 3x² - x²√(1 - x²)]/[x²(1 - x²)] dx
= ∫ [(3 - 3sin²θ - sin²θcosθ)(cosθ)]/(sin²θcos²θ) dθ
= ∫ (3 - 3sin²θ - sin²θcosθ)/(sin²θcosθ) dθ
= ∫ (3csc²θ - 3 - cosθ)/cosθ dθ
= ∫ [3(csc²θ - 1) - cosθ)]/cosθ dθ
= ∫ (3cot²θ - cosθ)/cosθ dθ
= ∫ (3cscθcotθ - 1) dθ
= - 3cscθ - θ + C
= - 3(1/x) - arcsin(x) + C
= - 3/x - arcsin(x) + C
提示：遇到√(1 ± x²)等等这种带根号的题目时，要换元三角函数来做

2、∫ (x⁴ + 2x² + x + 1)/[x(1 + x²)] dx
= ∫ [x + 1/x + 1/(x² + 1)] dx
= x²/2 + ln|x| + arctan(x) + C
提示：用多项式综合除法。