(a) Solve : \(2^{3y + 2} - 7(2^{2y + 2}) - 31(2^{y}) - 8 = 0, y \in R\). (b) Find \(\int (\sqrt{x^{2} + 1}) xdx\).
Question 1 Report
(a) Solve : \(2^{3y + 2} - 7(2^{2y + 2}) - 31(2^{y}) - 8 = 0, y \in R\).
(b) Find \(\int (\sqrt{x^{2} + 1}) xdx\).
Answer Details
(a) Let \(u=2^{y}\). Then \(2^{3y+2}=4u^{3}\) and \(2^{2y+2}=4u^{2}\), so
\[4u^{3}-7(4u^{2})-31u-8=0\Rightarrow 4u^{3}-28u^{2}-31u-8=0\]
Testing \(u=8\): \(4(512)-28(64)-31(8)-8=2048-1792-248-8=0\). So \((u-8)\) is a factor:
\[(u-8)(4u^{2}+4u+1)=0\Rightarrow (u-8)(2u+1)^{2}=0\]
Thus \(u=8\) or \(u=-\tfrac12\). Since \(u=2^{y}>0\), reject \(u=-\tfrac12\). Then \(2^{y}=8=2^{3}\), so
\[\boxed{y=3}\]
(b) \(\displaystyle\int x\sqrt{x^{2}+1}\,dx\). Let \(w=x^{2}+1\), so \(dw=2x\,dx\Rightarrow x\,dx=\tfrac12 dw\).
\[\int\sqrt{w}\cdot\tfrac12\,dw=\tfrac12\cdot\tfrac{2}{3}w^{3/2}+C=\tfrac13\left(x^{2}+1\right)^{3/2}+C\]
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