Question
Evaluate the following:
$\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{dx}$
$\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{dx}$ $=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}}\cdot\frac{1}{\text{x}^3}\text{dx}$
$=\int\sqrt{\frac{1+\text{x}^2}{\text{x}^2}}\cdot\frac{1}{\text{x}^3}\text{dx}$ $=\int\sqrt{\frac{1}{\text{x}^2}+1}\cdot\frac{1}{\text{x}^3}\text{dx}$
Put $1+\frac{1}{\text{x}^2}=\text{t}^2\Rightarrow\frac{-2}{\text{x}^3}\text{dx}=2\text{tdt}$
$\Rightarrow-\frac{1}{\text{x}^3}=\text{tdt}$
$\therefore\ \text{I}=-\int\text{t}^2\text{dt}=-\frac{\text{t}^3}{3}+\text{C}$ $=-\frac{1}{3}\Big(1+\frac{1}{\text{x}^2}\Big)^{\frac{3}{2}}+\text{c}$
$=\int\sqrt{\frac{1+\text{x}^2}{\text{x}^2}}\cdot\frac{1}{\text{x}^3}\text{dx}$ $=\int\sqrt{\frac{1}{\text{x}^2}+1}\cdot\frac{1}{\text{x}^3}\text{dx}$
Put $1+\frac{1}{\text{x}^2}=\text{t}^2\Rightarrow\frac{-2}{\text{x}^3}\text{dx}=2\text{tdt}$
$\Rightarrow-\frac{1}{\text{x}^3}=\text{tdt}$
$\therefore\ \text{I}=-\int\text{t}^2\text{dt}=-\frac{\text{t}^3}{3}+\text{C}$ $=-\frac{1}{3}\Big(1+\frac{1}{\text{x}^2}\Big)^{\frac{3}{2}}+\text{c}$
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