#shorts #indefinite Integral #find int 1/sqr root x²-6x+10 dx☺
∫ 10/(x² - 6x + 34) dx [-2, 8]. How to solve this integration problem?
Integral of 1/(x^2+6x+10) (substitution)
`int(dx)/(sqrt(x^(2)-6x+10))`
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微積分ヘルプ：dx/(x^2-6x-7)の積分 - 部分分数 - 積分技法
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Class 12 Maths Chapter 7 Example 9 | #dx/x^2+6x+13 | #dx/3x^2+13x-10 | #int(dx)/(sqrt(5x^(2)-2x))
Evaluateint (dx)/((x-3)^(3)sqrt(x^(2)-6x+10)). | 12 | INDEFINITE INTEGRAL | MATHS | ARIHANT MAT...
From the curve given in figure find `dy/dx at x=2, ` 6 and 10.
評価: int (dx)/(x^2-6x+13) | クラス 12 | 積分 | 数学 | ダウトナット
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