Pembahasan Kalkulus (Integral 2)


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


Penyelesaian :

Bentuk integral diatas dapat diubah menjadi
\begin{eqnarray*} {\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}{\left(x+1\right)^{2}+1}dx\\ & = & \ln|x^{2}+2x+2|-\tan^{-1}\left(x+1\right)+C \end{eqnarray*}
${\displaystyle \int\frac{x^{4}+8x^{2}+8}{x^{3}-4x}dx}$

Penyelesaian :


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

Bentuk ${\displaystyle \frac{12x^{2}+8}{x\left(x+2\right)\left(x-2\right)}}$
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{eqnarray*}
A+B+C & = & 12\\
-2B+2C & = & 0\\
-4A & = & 8
\end{eqnarray*}
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{eqnarray*}
\int\frac{x^{4}+8x^{2}+8}{x^{3}-4x}dx & = & \frac{1}{2}x^{2}+\int\frac{12x^{2}+8}{x\left(x+2\right)\left(x-2\right)}dx\\
& = & \frac{1}{2}x^{2}+\int\left(\frac{-2}{x}+\frac{7}{\left(x+2\right)}+\frac{7}{\left(x-2\right)}\right)dx\\
& = & \frac{1}{2}x^{2}+\int\frac{-2}{x}dx+\int\frac{7}{\left(x+2\right)}dx+\int\frac{7}{\left(x-2\right)}dx\\
& = & \frac{1}{2}x^{2}-2\int\frac{1}{x}dx+7\int\frac{1}{\left(x+2\right)}dx+7\int\frac{1}{\left(x-2\right)}dx\\
& = & \frac{1}{2}x^{2}-2\ln|x|+7\ln|x+2|+7\ln|x-2|+C
\end{eqnarray*}
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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