Evaluate the integral _ 0 ^ 1 x x^ 2 +1 d x using substitution. - Vidyadip

Question

Evaluate the integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$ using substitution.

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Answer

Given integral is: $\int_{0}^{1} \frac{x}{x^{2}+1} d x$
Let x² + 1 = t
⇒ 2xdx = dt
⇒ xdx = $\frac{1}{2}$ dt
When x = 0, t = 1 and when x = 1, t = 2
$\Rightarrow \int_{0}^{1} \frac{x}{x^{2}+1} d x=\int_{1}^{2} \frac{d t}{2 t}$
$=\frac{1}{2} \int_{1}^{2} \frac{d t}{t}$
$=\frac{1}{2}\left[\log |t| ]_{1}^{2}\right.$
= $\frac{1}{2}[\log 2-\log 1]$
= $\frac{1}{2} \log 2$

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