WAEC 2018 · Paper 2 · Q10

Worked solution (try it first)

Write each power in terms of $2^y$ : $2^{3y + 2} = 4(2^y)^3$ and $2^{2y + 2} = 4(2^y)^2$ .
Let $x = 2^y$ : $4x^3 - 28x^2 - 31x - 8 = 0$ .
Try factors of 8: $x = 8$ gives $2048 - 1792 - 248 - 8 = 0$ , so $(x - 8)$ is a factor.
Divide it out: $4x^3 - 28x^2 - 31x - 8 = (x - 8)(4x^2 + 4x + 1)$
$= (x - 8)(2x + 1)^2$ .
So $x = 8$ or $x = -\frac12$ .
But $x = 2^y$ is always positive, so reject $-\frac12$ .
$2^y = 8 = 2^3$ , so $y = 3$ .

Let $u = x^2 + 1$ , so $\dfrac{du}{dx} = 2x$ and $x\,dx = \frac12\,du$ .
The integral becomes $\displaystyle\int u^{\frac12} \times \frac12\,du = \frac12 \times \frac{u^{\frac32}}{\frac32} + c$
$= \frac13u^{\frac32} + c$ .
Put $u$ back: $\dfrac13(x^2 + 1)^{\frac32} + c$ .