微積分ヘルプ: 積分 ∫ √(25-16x^2 )/x dx - 三角関数の置換による積分

Integral of 1/sqrt (25 - 16x^2) dx-subtitution trigonometry

`int (dx)/(25-16x^(2))` is equal to-

Calculus Help: Integral of ∫ √(16x^2+25) dx - Integration by trigonometric substitution

Integration of 1/√16x²+25

`int(dx)/((16x^(2)-25))`

Integrate the function ∫ 1/root over 16x^2+25 dx

Calculus Help: Integral of dx/√(25-16x^2 ) - Integration by trigonometric substitution

微積分ヘルプ: ∫ 1/√(16x^2+25) dx - 三角関数の置換による積分 - テクニック

Calculus Help: Integral of (x^3dx)/√(16x^2-25) - Integration by substitution

Integration of 1/(16x^2 + 25) dx

integral of dx/16+25x^2

`int (dx)/(sqrt(16x^(2)+25))`

Integral of 1/sqrt(25-x^2)

Evaluate the following integrals: ∫√(16 x^2+25) d x

Integrate | 1/(sqrt(25-x^2)) dx

Evaluate the Integral 1/(25- 16y^2) dy. Partial Fraction Decomposition. Integrate

Calculus Help: ∫ dx/(x√(16x^2-1)) - Integration by trigonometric substitution - Techniques

.integration of sqrt16x^2+25

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