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∫1/1+sinxdx∫x4-x2/1+x2d
题目详情
∫1/1+sinxdx ∫x4-x2/1+x2d
▼优质解答
答案和解析
∫ dx/(1 + sinx)
= ∫ (1 - sinx)/[(1 + sinx)(1 - sinx)] dx
= ∫ (1 - sinx)/cos²x dx
= ∫ sec²x dx - ∫ secxtanx dx
= tanx - secx + C
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x⁴ - x² = x²(x² - 1)
= x²[(x² + 1) - 2]
= x²(x² + 1) - 2[(x² + 1) - 1]
= x²(x² + 1) - 2(x² + 1) + 2
∫ (x⁴ - x²)/(1 + x²) dx
= ∫ [x²(x² + 1) - 2(x² + 1) + 2]/(1 + x²) dx
= ∫ x² dx - 2∫ dx + 2∫ dx/(1 + x²)
= x³/3 - 2x + 2arctan(x) + C
= ∫ (1 - sinx)/[(1 + sinx)(1 - sinx)] dx
= ∫ (1 - sinx)/cos²x dx
= ∫ sec²x dx - ∫ secxtanx dx
= tanx - secx + C
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x⁴ - x² = x²(x² - 1)
= x²[(x² + 1) - 2]
= x²(x² + 1) - 2[(x² + 1) - 1]
= x²(x² + 1) - 2(x² + 1) + 2
∫ (x⁴ - x²)/(1 + x²) dx
= ∫ [x²(x² + 1) - 2(x² + 1) + 2]/(1 + x²) dx
= ∫ x² dx - 2∫ dx + 2∫ dx/(1 + x²)
= x³/3 - 2x + 2arctan(x) + C
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