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

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

Toolbox:
• (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,

$10xdx=dt$

$5 \times 2xdx=dt$

=>$2xdx=dt/5$

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 $I_2= \int\limits_0^1\large\frac{3dx}{5x^2+1}$

$=\frac{3}{5}\int\limits_0^1\large\frac{dx}{x^2+1/5}$

$=\frac{3}{5}\int\limits_0^1\large\frac{dx}{x^2+(\frac{1}{\sqrt 5})^2}$

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
edited Apr 11, 2016 by meena.p