Definite Integration — Maths STD 12 Science — Question

(c) : Let $I=\int_0^4 \frac{1}{1+\sqrt{x}} d x$
Put $1+\sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t$
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\Rightarrow \quad d x=2 \sqrt{x} d t=2(t-1) d t
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When $x=0, t=1$ and when $x=4, t=3$
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\begin{aligned}
& \therefore \quad I=\int_1^3 \frac{2(t-1)}{t} \cdot d t=\int_1^3 2 \cdot d t-\int_1^3 \frac{2}{t} \cdot d t=2[t]_1^3-2[\log t]_1^3 \\
& =2 \times 2-2(\log 3-\log 1)=4-2 \log 3=4 \log e-\log 3^2 \\
& =\log e^4-\log 9=\log \left(\frac{e^4}{9}\right) \quad \therefore \quad I=\log \left(\frac{e^4}{9}\right)
\end{aligned}
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