Question

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

Answer 1

Answer: 7

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