∫041/(1+√x)dx= ________

Correct answer: (C) log⁡(e^4)/9. Let I=∫04dx/(1+√x) put x=t^2 dx=2tdt Then, I=∫022tdt/(1+t)=2∫02(1+t-1)/(1+t)dt I=2∫021-1/(1+t)dt I=22-ln3=log(e^4)/9

Q89 / 150 · MHT CET 2019 (02 May Shift 1) · Mathematics

$\int_{0}^{4} \frac{1}{1 + \sqrt{x}} d x =$ ________

Let

I = \int_{0}^{4} \frac{d x}{1 + \sqrt{x}}

put

x = t^{2}

d x = 2 t d t

Then,

I = \int_{0}^{2} \frac{2 t d t}{1 + t} = 2 \int_{0}^{2} \frac{(1 + t) - 1}{1 + t} d t

I = 2 \int_{0}^{2} (1 - \frac{1}{1 + t}) d t

I = 2 (2 - l n 3) = l o g (\frac{e^{4}}{9})

Hence, the correct answer is option: 'C'.