Question
Evaluate the definite integral:
$
\int_2^4 \frac{x}{x^2+1} d x
$
$
\int_2^4 \frac{x}{x^2+1} d x
$
✓
Answer
Put $x ^2+1= t \Rightarrow 2 xdx = dt \Rightarrow xdx =\frac{1}{2} d t$
When $x =2, t =2^2+1=5$ and when $x =4, t =4^2+1=17$
$
\begin{array}{l}
\therefore \int_2^4 \frac{x}{x^2+1} d x=\frac{1}{2} \int_5^{17} \frac{d t}{t}=\frac{1}{2}[\log |t|]_5^{17} \\
=\frac{1}{2}(\log 17-\log 5)=\frac{1}{2} \log \frac{17}{5}
\end{array}
$
When $x =2, t =2^2+1=5$ and when $x =4, t =4^2+1=17$
$
\begin{array}{l}
\therefore \int_2^4 \frac{x}{x^2+1} d x=\frac{1}{2} \int_5^{17} \frac{d t}{t}=\frac{1}{2}[\log |t|]_5^{17} \\
=\frac{1}{2}(\log 17-\log 5)=\frac{1}{2} \log \frac{17}{5}
\end{array}
$
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