Correct Answer - Option 3 : 4 - 2log 3
Concept:
\(\rm \int \frac{1}{x} dx = \log x + c\)
Calculation:
I = \(\rm \int_0^4 \frac{1}{1+ \sqrt x}dx\)
Let 1 + \(\rm \sqrt x\) = t ....(1)
Differentiating with respect to x, we get
\(\rm \Rightarrow (0+\frac{1}{2\sqrt x})dx = dt\)
\(\rm \Rightarrow dx = {2\sqrt x}dt\)
From equation (1), we get
\(\rm \sqrt x\) = t - 1
∴ dx = 2(t - 1)dt
Now,
I = \(\rm \int_1^3 \frac{2(t-1)}{t}dt \)
= \(\rm 2\int_1^3 \left(1-\frac{1}{t} \right )dt \)
= \(\rm 2\left[t - \log t \right ]_1^3\)
= 2 [(3 - log 3) - (1 - log 1)]
= 2(2 - log 3)
= 4 - 2log 3
