Pembahasan Kalkulus (Integral 2)

${\displaystyle \int \frac{2x+1}{x^2+2x+2}dx}$

Penyelesaian :

Bentuk integral diatas dapat diubah menjadi
$$\begin{array}{rcl}{\displaystyle \int \frac{2x+1}{x^2+2x+2}dx}& =& {\displaystyle \int \frac{2x+2-1}{x^2+2x+2}dx}\\ & =& {\displaystyle \int \frac{2x+2}{x^2+2x+2}dx-\int \frac{1}{x^2+2x+2}dx}\\ & =& \ln |x^2+2x+2|-\int \frac{1}{x^2+2x+1+1}dx\\ & =& \ln |x^2+2x+2|-\int \frac{1}{{(x+1)}^2+1}dx\\ & =& \ln |x^2+2x+2|-{\tan }^{-1}(x+1)+C\end{array}$$

${\displaystyle \int \frac{x^4+8x^2+8}{x^3-4x}dx}$

Penyelesaian :

Bentuk integral diatas dapat diubah menjadi
$$\begin{array}{rcl}{\displaystyle \int \frac{x^4+8x^2+8}{x^3-4x}dx}& =& \int (x+\frac{12x^2+8}{x^3-4x})dx\\ & =& \int xdx+\int \frac{12x^2+8}{x(x^2-4)}dx\\ & =& \frac{1}{2}{x}^2+\int \frac{12x^2+8}{x(x+2)(x-2)}dx\end{array}$$

Bentuk ${\displaystyle \frac{12x^2+8}{x(x+2)(x-2)}}$
dengan menggunakan kesamaan suku banyak, maka dapat kita modifikasi menjadi

\begin{eqnarray*}
\frac{12x^{2}+8}{x\left(x+2\right)\left(x-2\right)} & = & \frac{A}{x}+\frac{B}{\left(x+2\right)}+\frac{C}{\left(x-2\right)}\
\frac{12x^{2}+8}{x\left(x+2\right)\left(x-2\right)} & = & \frac{A\left(x+2\right)\left(x-2\right)+Bx\left(x-2\right)+Cx\left(x+2\right)}{x\left(x+2\right)\left(x-2\right)}\
\frac{12x^{2}+8}{x\left(x+2\right)\left(x-2\right)} & = & \frac{A\left(x^{2}-4\right)+B\left(x^{2}-2x\right)+C\left(x^{2}+2x\right)}{x\left(x+2\right)\left(x-2\right)}\
\frac{12x^{2}+8}{x\left(x+2\right)\left(x-2\right)} & = & \frac{Ax^{2}-4A+Bx^{2}-2Bx+Cx^{2}+2Cx}{x\left(x+2\right)\left(x-2\right)}\
12x^{2}+8 & = & Ax^{2}-4A+Bx^{2}-2Bx+Cx^{2}+2Cx\
12x^{2}+8 & = & x^{2}\left(A+B+C\right)+\left(-2B+2C\right)x+\left(-4A\right)
\end{eqnarray*}Sehingga diperoleh
$$\begin{array}{rcl}A+B+C& =& 12\\ -2B+2C& =& 0\\ -4A& =& 8\end{array}$$

Dari persamaan diatas diperoleh $A=-2,B=7$ dan $C=7$ Sehingga dapat dituliskan

\[

\frac{12x^{2}+8}{x\left(x+2\right)\left(x-2\right)}=\frac{-2}{x}+\frac{7}{\left(x+2\right)}+\frac{7}{\left(x-2\right)}
\]
Diperoleh
$$\begin{array}{rcl}\int \frac{x^4+8x^2+8}{x^3-4x}dx& =& \frac{1}{2}{x}^2+\int \frac{12x^2+8}{x(x+2)(x-2)}dx\\ & =& \frac{1}{2}{x}^2+\int (\frac{-2}{x}+\frac{7}{(x+2)}+\frac{7}{(x-2)})dx\\ & =& \frac{1}{2}{x}^2+\int \frac{-2}{x}dx+\int \frac{7}{(x+2)}dx+\int \frac{7}{(x-2)}dx\\ & =& \frac{1}{2}{x}^2-2\int \frac{1}{x}dx+7\int \frac{1}{(x+2)}dx+7\int \frac{1}{(x-2)}dx\\ & =& \frac{1}{2}{x}^2-2\ln |x|+7\ln |x+2|+7\ln |x-2|+C\end{array}$$

Disimpulkan bahwa ${\displaystyle \int \frac{x^4+8x^2+8}{x^3-4x}dx=\frac{1}{2}{x}^2-2\ln |x|+7\ln |x+2|+7\ln |x-2|+C}$

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