sqrt(9+( x)^2)x dx is equal to: (where C is an arbitrary constant and consider _e x = x) Option 1: 1/2xsqrt(1+( x)^2) + 9| x + sqrt(9+( x)^2)| + C Option 2: 1/2 xsqrt(9+( x)^2) + 9/2| x + sqrt(9+( x)^2)| + C Option 3: xsqrt(9+( x)^2) - 9/2| x + sqrt(9+( x)^2)| + C Option 4: xsqrt(9+( x)^2) - 9| x + sqrt(9+( x)^2)| + C

The answer is '1/2 xsqrt(9+( x)^2) + 9/2| x + sqrt(9+( x)^2)| + C'. View full solution:

CUET Mathematics - sqrt(9+( x)^2)x dx is equal to: (where C is an arbitrary constant and consider _e x = x) | PYQs + Solutions

CUET Mathematics 2026 22nd May Shift 1

Calculus

Medium

common

$\frac{1}{2}x\sqrt{1+(\log x)^2} + 9\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

$\frac{1}{2}\log x\sqrt{9+(\log x)^2} + \frac{9}{2}\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

$\log x\sqrt{9+(\log x)^2} - \frac{9}{2}\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

$\log x\sqrt{9+(\log x)^2} - 9\log\left|\log x + \sqrt{9+(\log x)^2}\right| + C$

✅ Correct Option: 2