is equal to : [2025]

Question

$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

Smart Achievers · Accepted Answer

(3)

Let $f (x) = \frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}$

$= \frac{\sqrt{3 + x^{2}} - \sqrt{1 + x^{2}}}{3 + x^{2} - 1 - x^{2}} = \frac{\sqrt{3 + x^{2}} - \sqrt{1 + x^{2}}}{2}$

$∴ 4 \int_{0}^{1} f (x) d x = 4 \int_{0}^{1} \frac{\sqrt{3 + x^{2}} - \sqrt{1 + x^{2}}}{2} d x$

$= 2 [\int_{0}^{1} \sqrt{3 + x^{2}} d x - \int_{0}^{1} \sqrt{1 + x^{2}} d x]$

$= 2 {[\frac{x}{2} \sqrt{3 + x^{2}} + \frac{3}{2} \log_{e} | x + \sqrt{3 + x^{2}} | - \frac{x}{2} \sqrt{1 + x^{2}} - \frac{1}{2} \log_{e} | x + \sqrt{1 + x^{2}} |]}_{0}^{1}$

$= 2 + 3 \log_{e} 3 - \sqrt{2} - \log_{e} (1 + \sqrt{2}) - 3 \log_{e} \sqrt{3}$

$= 2 + 3 \log_{e} 3 - \sqrt{2} - \log_{e} (1 + \sqrt{2}) - \frac{3}{2} \log_{} 3$

$= 2 - \sqrt{2} - \log_{e} (1 + \sqrt{2}) + \frac{3}{2} \log_{e} 3$

$= 2 - \sqrt{2} - \log_{e} (1 + \sqrt{2}) + 3 \log_{e} \sqrt{3}$

$∴ 4 \int_{0}^{1} f (x) d x - 3 \log_{e} \sqrt{3} = 2 - \sqrt{2} - \log_{e} (1 + \sqrt{2})$ .

Solution

Q.
$4 \int_{0}^{1} (\frac{1}{\sqrt{3 + x^{2}} + \sqrt{1 + x^{2}}}) d x - 3 \log_{e} (\sqrt{3})$ is equal to : [2025]

1 $2 + \sqrt{2} - \log_{e} (1 + \sqrt{2})$

2 $2 + \sqrt{2} + \log_{e} (1 + \sqrt{2})$

3 $2 - \sqrt{2} - \log_{e} (1 + \sqrt{2})$

4 $2 - \sqrt{2} + \log_{e} (1 + \sqrt{2})$

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Solution

Q. 4∫01(13+x2+1+x2)dx–3 loge (3) is equal to : [2025]

1 2+2–loge (1+2)

2 2+2+loge (1+2)

3 2–2–loge (1+2)