If _0^11(5+2x-2x^2)(1+e^(2-4x))dx=(1)/( ) _e(( +1)/( )), , 0, then ^4- ^4 is equal to :

Question

If $\int _{0}^{1}\frac{1}{(5+2x-2{x}^{2})(1+{e}^{(2-4x)})}dx=\frac{1}{\alpha }{\log }_{e}(\frac{\alpha +1}{\beta }),\alpha ,\beta >0$, then ${\alpha }^{4}-{\beta }^{4}$

is equal to :

Accepted Answer

The given integral is:

$I={\int }_{1}^{0}\frac{dx}{(5+2x-2{x}^{2})(1+{e}^{2-4x})}$ .............(i)

Perform a substitution, $x	o 1-x$. This gives:

$I={\int }_{1}^{0}\frac{dx}{(5+2(1-x)-2(1-x{)}^{2})(1+{e}^{2-4(1-x)})}$

Simplify the expression inside the integral:

$I={\int }_{1}^{0}\frac{dx}{(5+2-2x-2(1-2x+{x}^{2}))(1+{e}^{4x-2})}$ ............(ii)

Add the original integral (i) and the integral after substitution (ii):

$2I={\int }_{1}^{0}\frac{dx}{5+2x-2{x}^{2}}$

Factor the quadratic expression in the denominator:

$2I={\int }_{1}^{0}\frac{dx}{2(\frac{11}{4}-{(x-\frac{1}{2})}^{2})}$

To solve this integral, we can perform a change of variables using the substitution

$x-\frac{1}{2}=\frac{\sqrt{11}}{2}	an u$, then $dx=\frac{\sqrt{11}}{2}{\sec }^{2}udu$:

$I=\frac{1}{\sqrt{11}}\int \frac{{\sec }^{2}u}{1+{	an }^{2}u}du$

Using the identity ${\sec }^{2}u=1+{	an }^{2}u$:

$I=\frac{1}{\sqrt{11}}\int du=\frac{1}{\sqrt{11}}(u+C)$

Now we need to find the limits of the integral after the substitution. If $x=0$, then $u={	an }^{-1}(-\frac{1}{\sqrt{11}})$. If $x=1$, then $u={	an }^{-1}(\frac{1}{\sqrt{11}})$. So, the integral becomes:

$I=\frac{1}{\sqrt{11}}[{	an }^{-1}(\frac{1}{\sqrt{11}})-{	an }^{-1}(-\frac{1}{\sqrt{11}})]$

Using the properties of the arctangent function, we can rewrite the integral as:

$I=\frac{1}{\sqrt{11}}\ln (\frac{\sqrt{11}+1}{\sqrt{10}})$

From this result, we have $\alpha =\sqrt{11}$ and $\beta =\sqrt{10}$. Now, we can find ${\alpha }^{4}-{\beta }^{4}$:

${\alpha }^{4}-{\beta }^{4}=(11{)}^{2}-(10{)}^{2}=121-100=21$

Thus, ${\alpha }^{4}-{\beta }^{4}=21$.

Solutions · 1