The correct option is A 4√(x) x 4log(1+√(x))+c ∫1 √(x)/1+√(x)dx =∫ #160; 1+√( x ) 2√( x ) #160;/ 1+√( x ) #160; dx =∫ #160;dx 2∫ ...

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Integration by Substitution

Solve ∫1-√x...

Question

Solve $\int \frac{1 - \sqrt{x}}{1 + \sqrt{x}} d x$

3 \sqrt{x} + \frac{x}{2} - 3 log (1 + \sqrt{x}) + c

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3 \sqrt{x} + 3 log (1 + \sqrt{x}) - \frac{1}{2} x + c

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3 \sqrt{x} - \frac{1}{2} x - 3 log (1 + \sqrt{x}) + + c

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4 \sqrt{x} - x - 4 log (1 + \sqrt{x}) + c

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Solution

The correct option is A $4 \sqrt{x} - x - 4 log (1 + \sqrt{x}) + c$
$\int \frac{1 - \sqrt{x}}{1 + \sqrt{x}} d x$
$= \int \frac{1 + \sqrt{x} - 2 \sqrt{x}}{1 + \sqrt{x}} d x$
$= \int d x - 2 \int \frac{\sqrt{x}}{1 + \sqrt{x}} d x$
$= x - 2 \int \frac{1 + \sqrt{x} - 1}{1 + \sqrt{x}} d x$
$= x - 2 \int d x + 2 \int \frac{1}{1 + \sqrt{x}} d x$
$= x - 2 x + 2 \int \frac{1}{1 + \sqrt{x}} d x$
Put $1 + \sqrt{x} = t$
$x = (t - 1)^{2}$
$\Rightarrow d x = 2 (t - 1) d t$
So, $I = x - 2 x + 4 \int \frac{t - 1}{t} d t$

$= - x + 4 \int 1 - \frac{1}{t} d t$

$= - x + 4 [t - log t] + C$
$= - x + 4 + 4 \sqrt{x} - 4 log (1 + \sqrt{x}) + C$
$= 4 \sqrt{x} - x - 4 log (1 + \sqrt{x}) + c$ where $c = C + 4$

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