Browse Questions

# Evaluate the integral using substitution$\int\limits_0^2\frac{dx}{x+4-x^2}$

Toolbox:
• (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
• $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}log \frac{|a+x|}{|a-x|}+c$
Given $\int\limits_0^2\frac{dx}{x+4-x^2}$

Consider $-x^2+x+4$

$=-(x^2-x-4)$

This can be written as $-\bigg[(x-1/2)^2-\frac{1}{4}-4\bigg]$

$=-\bigg[(x-1/2)^2-\frac{17}{4}\bigg]$

Therefore $I=\int \limits_0^2 \frac{dx}{(\frac{\sqrt {17}}{2})^2-{(x-1/2)^2}}$

This is of the form $\int \large\frac{dx}{a^2-x^2}=\frac{1}{2a} log \frac{|a+x|}{|a-x|}+c$

In the place of x we have (x-1/2) and in the place of a we have $\frac{\sqrt {17}}{2}$

Hence on integrating,

Therefore $I=\bigg[\large\frac{1}{2.\frac{\sqrt {17}}{2}} log \frac{\bigg|\frac{\sqrt {17}}{2}+(x-1/2)\bigg|}{\bigg|\frac{\sqrt {17}}{2}-(x- 1/2)\bigg|}\bigg]_0^2$

$=\bigg[\large\frac{1}{\sqrt {17}} log \frac{\bigg|\frac{\sqrt {17}}{2}+\frac{(2x-1)}{2}\bigg|}{\bigg|\frac{\sqrt {17}}{2}-\frac{(2x-1)}{2}\bigg|}\bigg]_0^2$

Applying limits and simplifying we get,

$I=\bigg[\large\frac{1}{\sqrt {17}} log \frac{|\sqrt {17}+3|}{|\sqrt {17}-3|} \times \frac{|\sqrt {17}+1|}{|\sqrt {17}-1|} \bigg]$

$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{17+3+4\sqrt {17}}{17+3-4\sqrt {17}}\bigg]$

$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{20+4\sqrt {17}}{20-4\sqrt {17}}\bigg]$

$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{5+\sqrt {17}}{5-\sqrt {17}}\bigg]$

Multiplying and dividing by the conjugat $5+\sqrt {17}$

$I=\large\frac{1}{\sqrt {17}} log \bigg[\frac{5+\sqrt {17}}{5-\sqrt {17}} \times \frac{5+\sqrt {17}}{5+\sqrt {17}} \bigg]$

$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{25+17+10\sqrt {17}}{8}\bigg]$

$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{42+10\sqrt {17}}{8}\bigg]$

$I=\large\frac{1}{\sqrt {17}} log \bigg[\frac{21+5\sqrt {17}}{8}\bigg]$