微積分ヘルプ: dx/(x^2 √(9-16x^2 )) の積分 - 三角関数の置換による積分
微積分ヘルプ: ∫ dx/(x√(9-16x^2 )) - 三角関数の置換による積分 - テクニック
Integration of 1/9-16x² #integral #cbse #maths #jee #bseb #nda #ncert #shorts
`int(1)/(16x^(2)+9)dx` is equal to
微積分ヘルプ: 積分 ∫ 1/(x^2 √(16x^2-9)) dx - 三角関数の置換による積分
If `int(1)/(sqrt(9-16x^(2)))dx=alphasin^(-1)(betax)+c`, then `alpha+(1)/(beta)=`
How to Integrate 8x/(√(9-16x^4)) dx
if `int (1)/(sqrt(9-16x^(2)))dx=alpha sin ^(-1) (beta x) +c.` then ` alpha +(1)/(beta )=`
12年生 – 1/sqrt(9 - 25 x^2) dxの積分 | 積分 | チュートリアルポイント
`int(x^2)/(9+16 x^6)dx`
int 1/sqrt(9-16x^(2)) dx = a sin^(-1) (bx) +c の場合、4a+3b= | クラス 12 | MHT-CET 2019 質問...
∫2x/√(1 - x⁴) dx. MIT Integration Bee 2011, Question 3, Qualifying Exam.
∫(2•ln(x) + 1)e^(ln(x))^2 dx MIT Integration Bee 2023, Qualifying Exam, Question 9. #integrationbee
int dx/(x^2 * sqrt(16x^2 - 9))
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Math- Derivative of 8x^2 =16x; d(6^2)/dx=?
Scientist Modi Ji | Modi ji ka mathematics | Extra 2ab | Prem Sir
∫ln(x)/(1 + x²) dx [0, ∞] = ? Method 2: x = eᵘ. Solving an integration problem with logarithm.
