Pembahasan Integral SBMPTN 2018 Matematika Dasar kode 552

Soal yang Akan Dibahas
$ \int \left( \frac{x^4-1}{x^3 + x} \right)^2 dx = .... $
A). $ \frac{1}{3}x^3 + \frac{1}{x} - 2x + C \, $
B). $ \frac{1}{3}x^3 - \frac{1}{x} - 2x + C \, $
C). $ \frac{1}{3}x^3 + \frac{1}{x} + 2x + C \, $
D). $ \frac{1}{3}x^3 - \frac{1}{x} + x + C \, $
E). $ \frac{1}{3}x^3 - \frac{1}{x} - x + C $

$\spadesuit $ Konsep Dasar
*). Rumus dasar integral :
$ \int \, ax^n \, dx = \frac{a}{n+1} x^{n+1} + c $
*). Sifat pemfaktoran :
i). $ a^2-b^2 = (a+b)(a-b) $ dan $ a^4 - b^4 = (a^2+b^2)(a^2 - b^2) $
ii). $ ( a - b)^2 = a^2 + b^2 - 2ab $

$\clubsuit $ Pembahasan
*). Menentukan hasil integralnya :
$\begin{align} & \int \left( \frac{x^4-1}{x^3 + x} \right)^2 dx \\ & = \int \left( \frac{(x^2+1)(x^2-1)}{x(x^2+1)} \right)^2 dx \\ & = \int \left( \frac{x^2-1}{x} \right)^2 dx \\ & = \int \left( x - \frac{1}{x} \right)^2 dx \\ & = \int \left( x^2 + \frac{1}{x^2} - 2 \right) dx \\ & = \int \left( x^2 + x^{-2} - 2 \right) dx \\ & = \frac{1}{2+1}x^{2+1} + \frac{1}{-2+1} x^{-2+1} - 2x + C \\ & = \frac{1}{3}x^3 + \frac{1}{-1} x^{-1} - 2x + C \\ & = \frac{1}{3}x^3 - \frac{1}{x} - 2x + C \end{align} $
Jadi, hasil $ \int \left( \frac{x^4-1}{x^3 + x} \right)^2 dx = \frac{1}{3}x^3 - \frac{1}{x} - 2x + C . \, \heartsuit $