Question

The value of the integral $\underset{0}{\overset{1}{\int }}\frac{1}{(5+2x-2x^2)(1+e^{2-4x})}dx$ is

• A. $\frac{1}{2\sqrt{11}}\ln \left|\frac{\sqrt{7}+3}{\sqrt{7}-3}\right|$
• B. $\frac{1}{2\sqrt{11}}\ln \left|\frac{\sqrt{11}+1}{\sqrt{11}-1}\right|$
• C. $\frac{1}{2\sqrt{11}}\ln \left|\frac{\sqrt{11}-1}{\sqrt{11}+1}\right|$
• D. $\frac{1}{2\sqrt{7}}\ln \left|\frac{\sqrt{3}+\sqrt{5}}{2}\right|$

Answer 1

The correct option is B $\frac{1}{2\sqrt{11}}\ln \left|\frac{\sqrt{11}+1}{\sqrt{11}-1}\right|$

Let $I=\underset{0}{\overset{1}{\int }}\frac{1}{(5+2x-2x^2)(1+e^{2-4x})}dx\cdots \left(1\right)$
Applying property $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$I=\underset{0}{\overset{1}{\int }}\frac{1}{(5+2(1-x)-2(1-x{)}^2)(1+e^{2-4(1-x)})}dx$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{1}{(5+2x-2x^2)(1+e^{-2+4x})}dx$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{e^{2-4x}}{(5+2x-2x^2)(1+e^{2-4x})}dx\cdots \left(2\right)$

From $\left(1\right)+\left(2\right),$
$2I=\underset{0}{\overset{1}{\int }}\frac{1}{(5+2x-2x^2)}dx$
$\Rightarrow 2I=\frac{-1}{2}\underset{0}{\overset{1}{\int }}\frac{1}{(x^2-x-\frac{5}{2})}dx$
$\Rightarrow 2I=\frac{-1}{2}\underset{0}{\overset{1}{\int }}\frac{1}{{(x-\frac{1}{2})}^2-{\left(\frac{\sqrt{11}}{2}\right)}^2}dx$
$\Rightarrow I=-\frac{1}{4}\cdot \frac{1}{2\left(\sqrt{11}/2\right)}{\left[\ln \left|\frac{x-\frac{1}{2}-\frac{\sqrt{11}}{2}}{x-\frac{1}{2}+\frac{\sqrt{11}}{2}}\right|\right]}_0^1$
$\therefore I=\frac{1}{2\sqrt{11}}\ln \left|\frac{\sqrt{11}+1}{\sqrt{11}-1}\right|$