Explanation:
(D):Let,
\(I=\int_{0}^{1} \frac{1}{x+\sqrt{x}} d x\)
\(=\int_{0}^{1} \frac{1}{\sqrt{\mathrm{x}}}\left(\frac{1}{\sqrt{\mathrm{x}}+1}\right) \mathrm{dx}\)
Let, \(\sqrt{\mathrm{x}}+1=\mathrm{t}\)
\(\frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{dx}=\mathrm{dt}\)
When, \(\mathrm{x}=0, \mathrm{t}=1\)
\(\mathrm{x}=1, \mathrm{t}=2\)
\(\mathrm{I}=\int_{0}^{1} \frac{2 \mathrm{dt}}{\mathrm{t}}\)
\(\mathrm{I}=2[\log \mathrm{t}]_{1}^{2}\)
\(\mathrm{I}=2[\log 2-\log 1]\)
\(\mathrm{I}=2 \log 2-0\)
\(\mathrm{I}=2 \log 2\)
\(\mathrm{I}=\log 2^{2}\)
\(\mathrm{I}=\log 4\)