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Evaluate the definite integral\[\int\limits_0^1\frac{2x+3}{5x^2+1}dx\]

$\begin{array}{1 1}\large \frac{1}{5} \log 6+\large \frac{3}{\sqrt 5}\tan^{-1}(\sqrt 5) \\ \large \frac{1}{5} \log 3-\large \frac{5}{\sqrt 3}\tan^{-1}(\sqrt 5) \\ \large \frac{1}{3} \log 3+\large \frac{5}{\sqrt 3}\tan^{-1}(\sqrt 3) \\ \large \frac{1}{3} \log 3-\large \frac{5}{\sqrt 3}\tan^{-1}(\sqrt 3)\end{array} $

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  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \frac{1}{1+x^2}dx=\tan^{-1}+c$
  • (iii)Method of substitution $I= \int f(x)dx. let\;f(x)=t, then \;f'(x)dx=dt.\;Hence\; I=\int t.dt$
Given$ \int\limits_0^1\large\frac{2x+3}{5x^2+1}dx$
We can split the terms and write as
$I= \int\limits_0^1\large\frac{2x}{5x^2+1}+\int\limits_0^1\large\frac{3}{5x^2+1}dx$
Consider $I_1= \int\limits_0^1\large\frac{2x}{5x^2+1}$
Let $x^2+1=t$
On differentiating we get,
$5 \times 2xdx=dt$
But the limits change.when substituting for x.
when x=0,t=1
when x=1,t=6
Therefore $ I_1=\int \limits_1^6 \large\frac{dt/5}{t}=\frac{1}{5}\int \limits_1^6\frac{dt}{t}$
On integrating we get,
$\bigg[\frac{1}{5}log t\bigg]_1^6$
On applying the limits,
$\frac{1}{5}[log 6-log1]$
We know $loga-logb=log(a/b),$ similarly.
$\frac{1}{5} log(6/1)=\frac{1}{5} log 6$
Consider $I_2= \int\limits_0^1\large\frac{3dx}{5x^2+1}$
$=\frac{3}{5}\int\limits_0^1\large\frac{dx}{x^2+(\frac{1}{\sqrt 5})^2}$
This is of the form $ \int \large\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a})+c$
Therefore $I=\frac{3}{5}\int\limits_0^1\large\frac{dx}{x^2+(\frac{1}{\sqrt 5})^2)}=\bigg[\large\frac{3}{5} \times \frac{1}{\frac{1}{\sqrt 5}}$
$\tan ^{-1}(\frac{x}{\frac{1}{\sqrt 5}})\bigg]_0^1$
On applying limits,
$\bigg[\frac{3 \sqrt 5}{5} tan^{-1}(x\sqrt 5)\bigg]_0^1=\frac{3}{\sqrt 5}\tan ^{-1}(\sqrt 5)$
Therefore $I=I_1+I_2$
$\int \limits_0^1 \large\frac{2x+3}{5x^2+1}dx=\frac{1}{5} log 6+\frac{3}{\sqrt 5}\tan^{-1}(\sqrt 5)$
answered Feb 11, 2013 by meena.p
edited Apr 11 by meena.p
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