Evaluate the following integrals: _ 2 ^4 x x^2+1 dx - Vidyadip

Question

Evaluate the following integrals:
$\int_{2}^\limits{4}\frac{\text{x}}{\text{x}^2+1}\text{ dx}$

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Answer

Let $\text{x}^2=\text{t}$ Then, $2\text{x dx}=\text{dt}$
When $\text{x}=2,\text{ t}=4$ and $\text{x}=4,\text{ t}=16$
$\therefore\ \text{I}=\int_{2}^\limits{4}\frac{\text{x}}{\text{x}^2+1}\text{ dx}$
$\Rightarrow\text{I}=\int_{4}^\limits{16}\frac{1}{2}\frac{\text{dt}}{\text{t}+1}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\log(\text{t}+1)\Big]^{16}_4$
$\Rightarrow\text{I}=\frac{1}{2}\log17-\frac{1}{2}\log5$
$\Rightarrow\text{I}=\frac{1}{2}\log\frac{17}{5}$

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