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int frac{1}{ {x}^{3} + 1} : dx
adalah …

Hint :
1) x³ + 1 = (x + 1)(x² – x + 1)
2) Kalau gak bisa u sub, paksakan
3) Lengkapi kuadrat sempurna​

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KALKULUS – INTEGRAL TAK TENTU

displaystyle int frac{1}{x^3 + 1} : sf{dx}

= displaystyle int frac{1}{(x + 1)(x^2 -x + 1)} : sf{dx}

Misalkan :

frac{1}{x^3 + 1} = frac{A}{x + 1} + frac{Bx + C}{x^2 -x + 1}

maka :

1 = A(x² -x + 1) + (Bx + C)(x + 1)

A(x² -x + 1) + (Bx + C)(x + 1) = 1

Ax² -Ax + A + Bx² + Bx + Cx + C = 1

(A + B)x² + (B -A + C)x + (A + C) = 1

maka didapat :

A + B = 0 => B = -A

A + C = 1 => C = 1 -A

B -A + C = 0

-A -A + 1 -A = 0

-3A = -1

A = ⅓

B = -A = -⅓

C = 1 -A = ⅔

maka :

frac{1}{x^3 + 1} = frac{ frac{1}{3} }{x + 1} + frac{ - frac{1}{3} x + frac{2}{3} }{x^2 -x + 1}

frac{1}{x^3 + 1} = frac{1}{3} ( frac{ 1}{x + 1} - frac{ x - 2 }{x^2 -x + 1})

maka integralnya berubah menjadi :

frac{1}{3} displaystyle int ( frac{ 1}{x + 1} - frac{ x - 2 }{x^2 -x + 1}) : sf{dx}

= frac{1}{3} displaystyle int frac{ 1}{x + 1} : sf{dx} - frac{1}{3} displaystyle int frac{ x - 2 }{x^2 -x + 1} : sf{dx}

misal b = x + 1 => db = dx

∫ 1/b db = ln|b| = ln|x + 1|

= frac{1}{3} ln | x + 1 | - frac{1}{3} displaystyle int frac{ x - 2 }{x^2 -x + 1} : sf{dx}

Nah, ok kita fokus ke integral displaystyle int frac{ x - 2 }{x^2 -x + 1} : sf{dx}

= displaystyle int frac{ x - 2 }{x^2 -x + frac{1}{4} + frac{3}{4} } : sf{dx}

= displaystyle int frac{ x - 2 }{(x - frac{1}{2} ) ^2 + ( frac{ sqrt{3}}{2} )^2 } : sf{dx}

misal u = x -½ => du = dx

x = u + ½ => x -2 = u -3/2

= displaystyle int frac{ u - frac{3}{2} }{ u^2 + ( frac{ sqrt{3}}{2} )^2 } : sf{du}

= displaystyle int frac{ u }{ u^2 + frac{3}{4} } : sf{dx} - frac{3}{2} displaystyle int frac{ 1 }{ u^2 + ( frac{ sqrt{3}}{2} )^2 } : sf{du}

nah, biar gak lupa bagian ini jangan dipisah

misalkan w = u² + ¾ => dw = 2u du

u du = ½ dw

= frac{1}{2} displaystyle int frac{ 1 }{w} : sf{dw} - frac{3}{2} displaystyle int frac{ 1 }{ u^2 + ( frac{ sqrt{3}}{2} )^2 } : sf{du}

= frac{1}{2} ln|w| - frac{3}{2} displaystyle int frac{ 1 }{ u^2 + ( frac{ sqrt{3}}{2} )^2 } : sf{du}

= frac{1}{2} ln (u^2 + frac{3}{4}) - frac{3}{2} displaystyle int frac{ 1 }{ u^2 + ( frac{ sqrt{3}}{2} )^2 } : sf{du}

= frac{1}{2} ln (x^2 -x + 1) - frac{3}{2} displaystyle int frac{ 1 }{ u^2 + frac{3}{4} } : sf{du}

misal u = ½√3 tan(θ) => θ = arctan(⅔u√3)

du = ½√3 sec²(θ) dθ

= frac{1}{2} ln (x^2 -x + 1) - frac{3}{2} displaystyle int frac{ 1 }{ (frac{sqrt{3}}{2} tan( theta) )^2 + frac{3}{4} } : ( frac{ sqrt{3}}{2} sec^2 ( theta) : sf{d} theta )

= frac{1}{2} ln (x^2 -x + 1) - frac{3 sqrt{3}}{4} displaystyle int frac{ 1 }{ frac{3}{4} (tan^2 ( theta) + 1) } : ( sec^2 ( theta) : sf{d} theta )

= frac{1}{2} ln (x^2 -x + 1) - frac{3 sqrt{3}}{4} displaystyle int frac{ 1 }{ frac{3}{4} sec ^2 ( theta )} : ( sec^2 ( theta) : sf{d} theta )

= frac{1}{2} ln (x^2 -x + 1) - sqrt{3} displaystyle int sf{d} theta

= frac{1}{2} ln (x^2 -x + 1) - sqrt{3} theta

= frac{1}{2} ln (x^2 -x + 1) - sqrt{3} arctan ( frac{2u sqrt{3}}{3} )

= frac{1}{2} ln (x^2 -x + 1) - sqrt{3} arctan ( frac{2 (x - frac{1}{2} ) sqrt{3}}{3} )

= frac{1}{2} ln (x^2 -x + 1) - sqrt{3} arctan ( frac{(2x sqrt{3} - sqrt{3}}{3} )

maka :

displaystyle int frac{ x - 2 }{x^2 -x + 1} : sf{dx} =frac{1}{2} ln (x^2 -x + 1) - sqrt{3} arctan ( frac{(2x sqrt{3} - sqrt{3}}{3} ) + C

maka :

displaystyle int frac{1}{x^3 + 1} = frac{1}{3} ln | x + 1 | - frac{1}{3} displaystyle int frac{ x - 2 }{x^2 -x + 1} : sf{dx}

displaystyle int frac{1}{x^3 + 1} = frac{1}{3} ln | x + 1 | - frac{1}{6} ln (x^2 -x + 1) + frac{ sqrt{3}}{3} arctan ( frac{(2x sqrt{3} - sqrt{3}}{3} ) + C

large boxed{boxed{displaystyle int frac{1}{x^3 + 1} = frac{1}{3} ln | x + 1 | - frac{1}{6} ln (x^2 -x + 1) + frac{ sqrt{3}}{3} arctan ( frac{(2x sqrt{3} - sqrt{3}}{3} ) + C }}