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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the integral using substitution\[\int\limits_0^1\frac{x}{x^2+1}dx\]

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  • (i)$\int \limits_a^b f(x)dx=F(b)-f(a)$
  • (ii)$\int f(x)dx,if f(x)=t\;then f'(x)dx=dt\; hence \int f(x)dx=\int t.dt$
  • $\int \frac{dx}{x}=log x+c$
Let $x^2+1=t$
Differentiating with respect to t,
Therefore $ xdx=\large\frac{dt}{2}$
substituting for t and dt,
Therefore $ \int \limits_0^1 \large\frac{dt/2}{t}=\frac{1}{2}\int \limits_1^2 \frac {dt}{t}$
$=\frac{1}{2}[log t]_1^2$
Applying the limits,
$=\frac{1}{2}[log 2-log1]$
$\int \limits_0^1 \large\frac{x}{x^2+1}dx=\frac{1}{2} log(2/1)=\frac{1}{2} log 2$



answered Feb 12, 2013 by meena.p
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