Question
Evaluate the following integrals:$\int\frac{1}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-4}}\text{ dx}$
✓
Answer
$\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\Big(\text{x}^{\frac{1}{3}}\Big)^2-2^2}}$
Let $\text{x}^{\frac{1}{3}}=\text{t}$
$\Rightarrow\frac{1}{3}\text{x}^{\frac{-2}{3}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{3\text{x}^{\frac{2}{3}}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}^{\frac{2}{3}}}=3\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=3\int\frac{\text{dt}}{\sqrt{\text{t}^2-2^2}}$
$=3\log\Big|\text{t}+\sqrt{\text{t}^2-2^2}\Big|+\text{C}$
$=3\log\Bigg|\text{x}^{\frac{1}{3}}+\sqrt{\text{x}^{\frac{2}{3}}-4}\Bigg|+\text{C}$
$=\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\Big(\text{x}^{\frac{1}{3}}\Big)^2-2^2}}$
Let $\text{x}^{\frac{1}{3}}=\text{t}$
$\Rightarrow\frac{1}{3}\text{x}^{\frac{-2}{3}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{3\text{x}^{\frac{2}{3}}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}^{\frac{2}{3}}}=3\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=3\int\frac{\text{dt}}{\sqrt{\text{t}^2-2^2}}$
$=3\log\Big|\text{t}+\sqrt{\text{t}^2-2^2}\Big|+\text{C}$
$=3\log\Bigg|\text{x}^{\frac{1}{3}}+\sqrt{\text{x}^{\frac{2}{3}}-4}\Bigg|+\text{C}$
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