If _0^1 1 (5+2 x -2 x ^2 ) (1+ e ^ (2-4 x) ) dx =1 _ e ( +1 ) , >… - Vidyadip

MCQ

If $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ $\alpha, \beta > 0$, then $\alpha^4-\beta^4$ is equal to:

✓
$21$
B
$0$
C
$19$
D
$-21$

✓

Answer

Correct option: A.

$21$

a
$I=\int \limits_0^1 \frac{d x}{\left(5+2 x-2 x^2\right)\left(1+ e ^{2-4 \pi}\right)}$

$x \rightarrow 1-x$

$I=\int \limits_0^1 \frac{e^{2-4 x} d x}{\left(5+2 x-2 x^2\right)\left(1+ e ^{2-4 x}\right)}$

Add $(i)$ and $(ii)$

$2 I=\int \limits_0^1 \frac{d x}{5+2 x-2 x^2}=\int \limits_0^1 \frac{d x}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)}$

$I=\frac{1}{\sqrt{11}} \ln \left(\frac{\sqrt{11}+1}{\sqrt{10}}\right) \quad \begin{array}{l}\alpha=\sqrt{11} \\\beta=\sqrt{10}\end{array}$

$\alpha^4-\beta^4=121-100=21$

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