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

Question

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

is equal to :

Accepted Answer

The given integral is:

$$
I=\int_0^1 \frac{dx}{\left(5+2x-2x^2\right)\left(1+e^{2-4x}\right)}
$$ .............(i)

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

$$
I=\int_0^1 \frac{dx}{\left(5+2(1-x)-2(1-x)^2\right)\left(1+e^{2-4(1-x)}\right)}
$$

Simplify the expression inside the integral:

$$
I=\int_0^1 \frac{dx}{\left(5+2-2x-2(1-2x+x^2)\right)\left(1+e^{4x -2}\right)}
$$ ............(ii)

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

$$
2I=\int_0^1 \frac{dx}{5+2x-2x^2}
$$

Factor the quadratic expression in the denominator:

$$
2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)}
$$

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

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

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

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

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

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

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

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

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

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