Question

If $\int \frac{4+x^{12}}{\left(x^{6}-2 x^{3}+2\right)} d x=\frac{x^{\alpha}}{\alpha}+\frac{x^{\beta}}{\gamma}+\delta x+ c$, then the last digit of the number $(\alpha-\beta+\gamma-\delta)^{99}$ is equal to (where ' $c$ ' is integration constant)

Answer 1

$\int \frac{\left(x^{6}-2 x^{3}+2\right)\left(x^{6}+2 x^{3}+2\right)}{\left(x^{6}-2 x^{3}+2\right)} d x$
$=\int\left(x^{6}+2 x^{3}+2\right) d x=\frac{x^{7}}{7}+\frac{x^{4}}{2}+2 x +c$
$\Rightarrow \alpha=7, \beta=4, \gamma=2, \delta=2$
$\Rightarrow(\alpha-\beta+\gamma-\delta)^{99}=(7-4+2-2)^{99}=3^{99}$
By cyclicity concept, $3^{99}$ has same last digit as $3^{3}$
$\therefore $ last digit $=7$.