Question
Prove the following Exercise:
$\int^{3}\limits_{1}\frac{\text{dx}}{\text{x}^{2}(\text{x}+1)}=\frac{2}{3}+\log\frac{2}{3}$
$\int^{3}\limits_{1}\frac{\text{dx}}{\text{x}^{2}(\text{x}+1)}=\frac{2}{3}+\log\frac{2}{3}$
✓
Answer
$\text{Let I}=\int^{3}\limits_{1}\frac{\text{dx}}{\text{x}^{2}(\text{x}+1)}$
Also, $\text{Let I}\int^{3}\limits_{1}\frac{1}{\text{x}^{2}(\text{x}+1)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}^{2}}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow1=\text{Ax(x}+1)+\text{B}\text{(x}+1)+\text{C}(\text{x}^{2})$
$\Rightarrow1=\text{Ax}^{2}+\text{Ax}+\text{Bx}+\text{B}+\text{Cx}^{2}$
Equating the Coefficient of $\text{x}^{2},\text{x}$ and constant term, we obtain
$\text{A}+\text{C}=0$
$\text{A}+\text{B}=0$
$\text{B}=1$
On solving these equation, we obtain
$\text{A}=-1,\text{C}=1,\ \text{and B}=1$
$\therefore\frac{1}{\text{x}^{2}(\text{x}+1)}=-\frac{1}{\text{x}}+\frac{1}{\text{x}^{2}}+\frac{1}{(\text{x}+1)}$
$\Rightarrow\text{I}=\int^{3}\limits_{1}\left\{-\frac{1}{\text{x}}+\frac{1}{\text{x}^{2}}+\frac{1}{\text{(x}+1)}\right\}\text{dx}$
$=\bigg[-\log\text{x}-\frac{1}{\text{x}}+\log(\text{x}+1)\bigg]^{3}_{1}$
$=\bigg[-\log\bigg(\frac{\text{x}+1}{\text{x}}\bigg)-\frac{1}{\text{x}}\bigg]^{3}_{1}$
$=\log4-\log3-\log2+\frac{2}{3}$
$=-\log2-\log3+\frac{2}{3}$
$=\log\Big(\frac{2}{3}\Big)+\frac{2}{3}$
Hence, the given result is proved
Also, $\text{Let I}\int^{3}\limits_{1}\frac{1}{\text{x}^{2}(\text{x}+1)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}^{2}}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow1=\text{Ax(x}+1)+\text{B}\text{(x}+1)+\text{C}(\text{x}^{2})$
$\Rightarrow1=\text{Ax}^{2}+\text{Ax}+\text{Bx}+\text{B}+\text{Cx}^{2}$
Equating the Coefficient of $\text{x}^{2},\text{x}$ and constant term, we obtain
$\text{A}+\text{C}=0$
$\text{A}+\text{B}=0$
$\text{B}=1$
On solving these equation, we obtain
$\text{A}=-1,\text{C}=1,\ \text{and B}=1$
$\therefore\frac{1}{\text{x}^{2}(\text{x}+1)}=-\frac{1}{\text{x}}+\frac{1}{\text{x}^{2}}+\frac{1}{(\text{x}+1)}$
$\Rightarrow\text{I}=\int^{3}\limits_{1}\left\{-\frac{1}{\text{x}}+\frac{1}{\text{x}^{2}}+\frac{1}{\text{(x}+1)}\right\}\text{dx}$
$=\bigg[-\log\text{x}-\frac{1}{\text{x}}+\log(\text{x}+1)\bigg]^{3}_{1}$
$=\bigg[-\log\bigg(\frac{\text{x}+1}{\text{x}}\bigg)-\frac{1}{\text{x}}\bigg]^{3}_{1}$
$=\log4-\log3-\log2+\frac{2}{3}$
$=-\log2-\log3+\frac{2}{3}$
$=\log\Big(\frac{2}{3}\Big)+\frac{2}{3}$
Hence, the given result is proved
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