`int(1)/(16x^(2)+9)dx` is equal to

`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

微積分ヘルプ: 積分 ∫ dx/(4x√(16x^2-9)) - 三角関数の置換による積分

Integration of 1/16+9x^2 dx

integrate dx/(1 + 16x^2) dx

Integrate | 1/sqrt(16-x^2) dx

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

If `int(1)/(sqrt(9-16x^(2)))dx=alphasin^(-1)(betax)+c`, then `alpha+(1)/(beta)=`

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if `int (1)/(sqrt(9-16x^(2)))dx=alpha sin ^(-1) (beta x) +c.` then ` alpha +(1)/(beta )=`

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