_2^4 x x^2+1 d x - Vidyadip

Question

$\int_2^4 \frac{x}{x^2+1} d x$

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Answer

$
\text { Let } I =\int_2^4 \frac{x}{x^2+1} \cdot d x
$
Put $x^2+1=t$
$
\begin{aligned}
& \therefore 2 x \cdot dx = dt \\
& \therefore x \cdot dx =\frac{ dt }{2}
\end{aligned}
$
When $x=2, t=2^2+1=5$
When $x=4, t=42+1=17$
$
\begin{aligned}
& \therefore I =\int_5^{17} \frac{1}{ t } \cdot \frac{ dt }{2} \\
& =\frac{1}{2} \int_5^{17} \frac{ dt }{ t } \\
& =\frac{1}{2}[\log | t |]_5^{17} \\
& =\frac{1}{2}(\log 17-\log 5) \\
& \therefore I =\frac{1}{2} \log \left(\frac{17}{5}\right) .
\end{aligned}
$

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