is equal to : [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

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

Ans.
(3)

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

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

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

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

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

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

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

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

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

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

Want Us to Email you About Special Offers & Updates?