Let I= _0^9 1 1+x d x Put x=t, i.e. x=t^2 & d x=2 t d t & When x=0, t=0 & When x=9, t=3 ^3 1 & =2 _0^3 [(1+t)-1 1+t ] d t & =2 _0^3 (1-1 1+t ) d t & =2[t- |1+t|]_0^3 & =2[(3- 4)-0] & =6-2 4=6-4 2 .

_0^9 1 1+x d x

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$\int_0^9 \frac{1}{1+\sqrt{x}} d x$

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Answer

Let $I=\int_0^9 \frac{1}{1+\sqrt{x}} d x$
Put $\sqrt{x}=t$, i.e. $x=t^2$
$
\begin{aligned}
& \therefore d x=2 t d t \\
& \text { When } x=0, t=0 \\
& \text { When } x=9, t=3
\end{aligned}
$
$
\therefore{ }^3 1
$
$
\begin{aligned}
& =2 \int_0^3\left[\frac{(1+t)-1}{1+t}\right] d t \\
& =2 \int_0^3\left(1-\frac{1}{1+t}\right) d t \\
& =2[t-\log |1+t|]_0^3 \\
& =2[(3-\log 4)-0] \\
& =6-2 \log 4=6-4 \log 2 .
\end{aligned}
$

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