(a) Solve : \(2^{3y + 2} – 7(2^{2y + 2}) – 31(2^{y}) – 8 = 0, y \in R\).
(b) Find \(\int (\sqrt{x^{2} + 1}) xdx\).
Explanation
(a) \(2^{3y + 2} - 7(2^{2y + 2}) - 31(2^{y}) - 8 = 0\)
\(\equiv (2^{2})(2^{y})^{3} - 7(2^{2})(2^{y})^{2} - 31(2^{y}) - 8 = 0\)
Let \(2^{y} = x\),
\(4x^{3} - 28x^{2} - 31x - 8 = 0\)
If x = 8, \(4(8^{3}) - 28(8^{2}) - 31(8) - 8 = 2048 - 1792 - 248 - 8\)
= \(0\)
\(\therefore (x - 8) \text{is a factor}\)
Dividing \(\frac{4x^{3} - 28x^{2} - 31x - 8}{x - 8}\), we get
= \(4x^{2} + 4x + 1 \)
= \((2x + 1)^{2}\)
Therefore, \(x = -\frac{1}{2}, -\frac{1}{2}, 8\)
Recall, \(x = 2^{y}\)
\(2^{y} = -\frac{1}{2}\) has no real solution.
\(2^{y} = 8 = 2^{3} \implies y = 3\)
(b) \(\int (\sqrt{x^{2} + 1}) xdx\)
Let \(u^{2} = x^{2} + 1; 2\frac{\mathrm d u}{\mathrm d x} = 2x\)
\(\mathrm d u = x \mathrm d x\)
\(\therefore \int (\sqrt{x^{2} + 1}) xdx = \int (\sqrt{u^{2}}) \mathrm d u\)
= \(\int u \mathrm d u\)
= \(\frac{u^{2}}{2} + c\)
= \(\frac{x^{2} + 1}{2} + c\)