Evaluate the following integrals: 1 x(x^6+1) dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\text{x}(\text{x}^6+1)}\text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{\text{x}(\text{x}^6+1)}\text{dx}$
$=\int\frac{\text{x}^5}{\text{x}^6(\text{x}^6+1)}\text{dx}$
Let $\text{x}^6=\text{t}$
$\Rightarrow6\text{x}^5\text{dx = dt}$
$\Rightarrow\text{x}^5\text{dx}=\frac{\text{dt}}{6}$
$\text{I}=\frac{1}{6}\int\frac{\text{dt}}{\text{t}(\text{t}+1)}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}^2+\text{t}}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}^2+2\text{t}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}$
$=\frac{1}{6}\int\frac{\text{dt}}{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}$
Let $\text{t}+\frac{1}{2}=\text{u}$
$\Rightarrow\text{dt = du}$
$\text{I}=\frac{1}{6}\int\frac{\text{du}}{\text{u}^2-\big(\frac{1}{2}\big)^2}$
$=\frac{1}{6}\times\frac{1}{2\big(\frac{1}{2}\big)}\log\Bigg|\frac{\text{u}-\frac{1}{2}}{\text{u}+\frac{1}{2}}\Bigg|+\text{C}$ $\bigg[\text{Since}\int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\bigg]$
$\text{I}=\frac{1}{6}\log\Bigg|\frac{\text{t}+\frac{1}{2}-\frac{1}{2}}{\text{t}+\frac{1}{2}+\frac{1}{2}}\Bigg|+\text{C}$
$\text{I}=\frac{1}{6}\log\bigg|\frac{\text{x}^6}{\text{x}^6+1}\bigg|+\text{C}$

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