Q29 | Evaluate ∫(x^3+5x^2-4)/x^2 dx | Integral of (x^3+5x^2-4)/x^2 | Integration of (x^3+5x^2-4)/x^2

Integrate (5x+1)/(2x²+4x+3) wrt x

Integral 2x+3/x^2+4x+5 dx

Integral dx / 5 - 2x² + 4x @EAG

微積分ヘルプ: ∫ (x+3)/√(5-4x-x^2 ) dx の積分 - 三角関数の置換による積分

評価: `int 1/sqrt(5-4x-2x^2)dx`

Integration (Calculus)

Integral x+5/x^2+4x+5 dx

Find the following integrals: (i) `int(x+2)/(2x^2+6x+5)dx` (ii) `int(x+3)/(sqrt(5-4x+x^2))dx`...

Calculus Help: Integral ∫ 2x(x^2+3)^5 dx - Integration by substitution - 置換による統合

∫(x³ + x² + x + 2)/(x⁴ + 3x² + 2) dx. Fast solution for indefinite integral without substitution.

Calculus Help: Integral ∫ x^2/√(5-4x^2 ) dx - Integration by trigonometric substitution

Differentiation and integration important formulas||integration formula

x=0 から 1 までの (2x^4 - 3x^2 + 5) dx の定積分を求めます。

∫1/√(x² + x) dx = ? Solve without using substitution method.

`int[x+3]/sqrt[5-4x-x^2].dx`

Evaluate the Integral (x^3 -2x^2 + x +1)/(x^4 + 5x^2 +4) dx. Partial Fraction Decomposition

微積分ヘルプ: 積分 ∫ (3x^2)/√(5-4x^3 ) dx, ∫ (x-4)/(x^2-8x+1) dx - 置換積分

Differentiation and Integration formula

Evaluate `int((x+3))/(sqrt(5-4x-x^(2)))dx`.