Evaluate the following intregals: 1 x(x^n+1) dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{1}{\text{x}(\text{x}^\text{n}+1)}\text{ dx}$

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Answer

$\frac{1}{\text{x}(\text{x}^\text{n}+1)}$
Multiplying numerator and denominatoe by $\text{x}^{\text{n}-1},$ we obtain
$\frac{1}{\text{x}(\text{x}^\text{n}+1)}=\frac{\text{x}^{\text{n}-1}}{\text{x}^{\text{n}-1}\text{x}(\text{x}^\text{x}+1)}=\frac{\text{x}^{\text{n}-1}}{\text{x}^{\text{n}}(\text{x}^\text{n}+1)}$
Let $\text{x}^\text{n}=\text{t}\Rightarrow\text{x}^{\text{n}-1}\text{dx}=\text{dt}$
$\therefore\int\frac{1}{\text{x}(\text{x}^\text{n}+1)}\ \text{dx}=\int\frac{\text{x}^{\text{n}-1}}{\text{x}^\text{n}(\text{x}^\text{n}+1)}\ \text{dx}=\frac{1}{\text{n}}\int\frac{1}{\text{t}(\text{t}+1)}\ \text{dt}$
Let $\frac{1}{\text{t}(\text{t}+1)}=\frac{\text{A}}{\text{t}}+\frac{\text{B}}{(\text{t}+1)}$
$\text{I}=\text{A}(1+\text{t})+\text{Bt}\ ...(1)$
Substituting t = 0, -1 in equation (1), we obtain
A = 1 and B = -1
$\therefore\frac{1}{\text{t}(\text{t}+1)}=\frac{1}{\text{t}}-\frac{1}{(1+\text{t})}$
$\Rightarrow\int\frac{1}{\text{x}(\text{x}^\text{n}+1)}\ \text{dx}=\frac{1}{\text{n}}\int\Big\{\frac{1}{\text{t}}-\frac{1}{(\text{t}+1)}\Big\}\text{dx}$
$=\frac{1}{\text{n}}[\log|\text{t}|-\log|\text{t}+1|]+\text{C}$
$=-\frac{1}{\text{n}}[\log|\text{x}^\text{n}|-\log|\text{x}^\text{n}+1|]\text{C}$
$=\frac{1}{\text{n}}\log\Big|\frac{\text{x}^\text{n}}{\text{x}^\text{n}+1}\Big|+\text{C}$

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