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