If \( \int_{0}^{4}\displaystyle\frac{dx}{\sqrt{x^2+2x+3}}=\log_e\left(\frac{\alpha+3\sqrt{\beta}}{1+\sqrt{\beta}}\right),\beta\gt 0 \), then \( \alpha^2+\beta^2 \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 34
Completing the square,
\( \displaystyle x^2+2x+3=(x+1)^2+(\sqrt{2})^2 \)
The standard integral
\( \displaystyle \int\frac{du}{\sqrt{u^2+a^2}}=\log\left|u+\sqrt{u^2+a^2}\right| \)
with u = x + 1 gives
\( \displaystyle \int_{0}^{4}\frac{dx}{\sqrt{(x+1)^2+2}}=\Big[\log\left|x+1+\sqrt{x^2+2x+3}\right|\Big]_{0}^{4} \)
\( \displaystyle =\log\left(5+\sqrt{27}\right)-\log\left(1+\sqrt{3}\right) \)
\( \displaystyle =\log_e\left(\frac{5+3\sqrt{3}}{1+\sqrt{3}}\right) \)
Comparing with \( \log_e\left(\frac{\alpha+3\sqrt{\beta}}{1+\sqrt{\beta}}\right) \): \( \sqrt{\beta}=\sqrt{3} \), so β = 3 and α = 5.
\( \displaystyle \alpha^2+\beta^2=25+9=34 \)