Let I= _2^3 x x^2+1 d x Put x^2+1=t 2 x d x=d t x d x=d t 2 When x=2, t=4+1=5 When x=3, t=9+1=10 I & = _5^ 10 1 t d t 2 =1 2 _5^ 10 1 t d t & =1 2 [ |t|]_5^ 10 & =1 2 ( 10- 5)=1 2 (10 5 ) & =1 2 2= 2 .

_2^3 x x^2+1 d x

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Question

$\int_2^3 \frac{x}{x^2+1} d x$

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Answer

Let $I=\int_2^3 \frac{x}{x^2+1} d x$
Put $x^2+1=t \quad \therefore 2 x d x=d t$
$
\therefore x d x=\frac{d t}{2}
$
When $x=2, t=4+1=5$
When $x=3, t=9+1=10$
$
\begin{aligned}
\therefore I & =\int_5^{10} \frac{1}{t} \cdot \frac{d t}{2}=\frac{1}{2} \int_5^{10} \frac{1}{t} d t \\
& =\frac{1}{2}[\log |t|]_5^{10}
\end{aligned}
$
$
\begin{aligned}
& =\frac{1}{2}(\log 10-\log 5)=\frac{1}{2} \log \left(\frac{10}{5}\right) \\
& =\frac{1}{2} \log 2=\log \sqrt{2} .
\end{aligned}
$

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