How to find the indefinite integral of [(x-2)/(x^2-4x+1)^2]dx

微積分ヘルプ: (x-3)/√(2x^2-4x-1) dx の積分 - 三角関数の置換による積分

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integral of dx/x2+4x+8 || integral of dx/x²+4x+8 || Integration of dx/x^2 +4x +8 || #term2

integrate (2x + 1)/((x ^ 2 + 4x + 1) ^ (3/2)) dx — Well Explained

微積分ヘルプ: (2x-1)/√(x^2+4x+5) dx の積分 - 置換積分

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Evaluate the limit (x^2/(1-x^2))/(4x -1 + e^(-4x)). Maclaurin series

integral of (x - 2)/sqrt(x^2 - 4x + 3) dx

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Evaluate: ∫▒〖(x^4-2x^2+4x+1)/(x^3-x^2-x+1) dx〗

Q44 Class 12 Maths Integration of (2x+1 )/sqrt(x^2+4x+3)dx Important Questions CBSE Board Exam 2019

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x^2 - 4x - 1 = 0