Evaluate the follwing intregals: 1 x(x^4-1) dx - Vidyadip

Question

Evaluate the follwing intregals:
$\int\frac{1}{\text{x}(\text{x}^4-1)}\ \text{dx}$

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Answer

We have,
$\text{I}=\int\frac{\text{dx}}{\text{x}(\text{x}^4-1)}$
$=\int\frac{\text{x}^3\text{dx}}{\text{x}^4(\text{x}^4-1)}$
Putting $\text{x}^4=\text{t}$
$\Rightarrow4\text{x}^3\text{dx}=\text{dt}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
$\therefore\text{I}=\frac{1}{4}\int\frac{\text{dt}}{\text{t}(\text{t}-1)}$
Let $\frac{1}{\text{t}(\text{t}-1)}=\frac{\text{A}}{\text{t}}+\frac{\text{B}}{\text{t}-1}$
$\rightarrow\frac{1}{\text{t}(\text{t}-1)}=\frac{\text{A}(\text{t}-1)+\text{B}\text{t}}{(\text{t}-1)}$
$\Rightarrow1=\text{A}(\text{t}-1)+\text{Bt}$
Putting t - 1 = 0
⇒ t = 1
$\therefore$ 1 = A × 0 + B (1)
⇒ B = 1
Putting t = 0
$\therefore$ 1 = A (0 - 1) + B × 0
⇒ A = -1
$\therefore$ $\text{I}=-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}+\frac{1}{4}\int\frac{\text{dt}}{\text{t}-1}$
$=-\frac{1}{4}\log|\text{t}|+\frac{1}{4}\log|\text{t}-1|+\text{C}$
$=\frac{1}{4}\log\Big|\frac{\text{t}-1}{\text{t}}\Big|+\text{C}$
$=\frac{1}{4}\log\Big|\frac{\text{x}^2-1}{\text{x}^4}\Big|+\text{C}$

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