Evaluate the following intregals: x^2+1 (2x+1)(x^2-1) dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}\text{ dx}$

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Answer

We have,$\text{I}=\int\frac{\text{x}^2+1}{(2\text{x}+1)(\text{x}^2-1)}$
$\text{I}=\int\frac{(\text{x}^2+1)\text{dx}}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}$
Let $\text{I}=\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}}{2\text{x}+1}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow\int\frac{(\text{x}^2+1)}{(2\text{x}+1)(\text{x}^2-1)(\text{x}+1)}=\frac{\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)+\text{C}(2\text{x}+1)(\text{x}-1)}{(2\text{x}-1)(\text{x}-1)(\text{x}+1)}$
$\Rightarrow\text{x}^2+1=\text{A}(\text{x}^2-1)+\text{B}(2\text{x}+1)(\text{x}+1)\\+\text{C}(2\text{x}+1)(\text{x}-1)$
Putting x - 1 = 0
⇒ x = 1
1 + 1 = A × 0 + B × 0 + C (-2 + 1) (-1 - 1)
⇒ 2 = B (3) (2)
$\Rightarrow\text{B}=\frac{1}{3}$
Putting x + 1 = 0
⇒ x = -1
1 + 1 = A × 0 + B (-2 + 1)(-1 - 1)
⇒ 2 = C (-1) (-2)
⇒ C = 1
Putting 2x + 1 = 0
$\Rightarrow\text{x}=-\frac{1}{2}$
$\Big(-\frac{1}{2}\Big)^2+1=\text{A}\Big(\frac{1}{4}-1\Big)$
$\Rightarrow\frac{1}{4}+1=\text{A}\Big(-\frac{3}{4}\Big)$
$\Rightarrow\frac{5}{4}=\text{A}\Big(-\frac{3}{4}\Big)$
$\text{A}=-\frac{5}{3}$
$\therefore\text{I}=-\frac{5}{3}\int\frac{\text{dx}}{2\text{x}+1}+\frac{1}{3}\int\frac{\text{dx}}{\text{x}-1}+\int\frac{\text{dx}}{\text{x}+1}$
$=-\frac{5}{3}\times\frac{\log|2\text{x}+1|}{2}+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$
$=-\frac{5}{6}\log|2\text{x}-1|+\frac{1}{3}\log|\text{x}-1|+\log|\text{x}+1|+\text{C}$

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