∫6x+5.sqrt 6-2x^2+x dx
Integral of sqrt(x^6 - 6x^5 + 9x^4 + 2x^3 - 6x^2 + 1)
Evaluate ∫(6x+5)root6-2x^2+x dx & ∫ (3x-2)root2x^2-x+1 dx 7M IMP (Q5,Q6)
Q23 | ∫(x^2+4x)/(x^3+6x^2+5) dx | Integral of (x^2+4x)/(x^3+6x^2+5)dx | Indefinite Integration
Integration by Perfect Square: How to Integrate (x+2)/(2x^2+6x+5) by Completing Square Method
Integral of (2x-3)/(2x^2-6x+5) (substitution)
∫1/√(x - x²) dx. MIT Integration Bee 2022, Question 18, Qualifying Exam.
Integration 7 Marks Important Question 11 | (6x+5)√(6-2x^2+x)
微積分ヘルプ: (x^3+2x)^5 (6x^2+4)dx の積分 - 置換積分
Trig Substitution: Integral x/sqrt(x^2-6x+8)
Calculus Help: Integral of ∫ (x+5)/(x^2-6x+13) dx - Integration by trigonometric substitution
∫x⁴/(3 - 6x + 6x² - 4x³ + 2x⁴) dx = ?? MIT Integration Bee 2024, Qualifying Exam, Question 19.
微積分ヘルプ: 積分 ∫ (x+1)/(9x^2+6x+5) dx - 三角関数の置換による積分
Find the following integrals: (i) `int(x+2)/(2x^2+6x+5)dx` (ii) `int(x+3)/(sqrt(5-4x+x^2))dx`...
`int(6x-5)/(sqrt(3x^2-5x+1))dx`
`int (x+2)/(2x^(2)+6x+5) dx` の値を求めます。
積分 x/sqrt(x^2-6x+8)
∫(cos⁴(x) - sin⁴(x)) dx. MIT Integration Bee 2015, Question 11, Qualifying Exam.
Integral of (x+1)/sqrt(x^2+2x+5) (substitution)
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