Evaluate the following integrals: x^3-1 x^3+x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^3-1}{\text{x}^3+\text{x}}\text{ dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}^3-1}{\text{x}^3+\text{x}}\text{ dx}$
$=\int1-\frac{(\text{x}+1)}{\text{x}^3+\text{x}}\ \text{dx}$
$=\int\text{dx}-\frac{\text{x}+1}{\text{x}^3+\text{x}}\ \text{dx}$
Let $\frac{\text{x}+1}{\text{x}(\text{x}^2+1)}=\frac{\text{A}}{\text{x}}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\text{x}+1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x})$
$=(\text{A}+\text{B})\text{x}^2+(\text{B}+\text{C})\text{x}+\text{A}$
Equating similar terms, we get,
$\text{A}+\text{B}=0,\text{C}=1,\text{A}=1$
Solving we get A = 1, B = -1, C =1
Thus,
$\text{I}=-\int\frac{\text{dx}}{\text{x}}-\int\frac{-\text{x}+1}{\text{x}^2+1}\text{dx}+\int\text{dx}$
$=-\log|\text{x}|+\int\frac{\text{xdx}}{\text{x}^2+1}-\int\frac{\text{dx}}{\text{x}^2+1}+\int\text{dx}$
$\therefore\text{I}=\text{x}-\log|\text{x}|+\frac{1}{2}\log|\text{x}^2+1|-\tan^{-1}\text{x}+\text{C}$
$\therefore\text{I}=\text{x}-\log|\text{x}|+\frac{1}{2}\log|\text{x}^2+1|-\tan^{-1}\text{x}+\text{C}$

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