Evaluate the integral: int sqrt{4x(4x + 4)} , | vedclass

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Question

(A) The given integral is $I = \int \sqrt{16x^2 + 16x} \, dx = 4 \int \sqrt{x^2 + x} \, dx$.Complete the square: $x^2 + x = (x + \frac{1}{2})^2 - (\fr

Vedclass · Accepted Answer

$\frac{1}{\log 2} [\frac{2^x}{2} \sqrt{4^x + 4} + 2 \log |2^x + \sqrt{4^x + 4}|] + c$

Vedclass · Answer

(A) The given integral is $I = \int \sqrt{16x^2 + 16x} \, dx = 4 \int \sqrt{x^2 + x} \, dx$.Complete the square: $x^2 + x = (x + \frac{1}{2})^2 - (\frac{1}{2})^2$.Using the formula $\int \sqrt{u^2 - a^2} \, du = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \log |u + \sqrt{u^2 - a^2}| + c$, where $u = x + \frac{1}{2}$ and $a = \frac{1}{2}$.$I = 4 [\frac{x + 1/2}{2} \sqrt{(x + 1/2)^2 - (1/2)^2} - \frac{1/4}{2} \log |(x + 1/2) + \sqrt{(x + 1/2)^2 - (1/2)^2}|] + c$.$I = 2(2x + 1) \sqrt{x^2 + x} - \frac{1}{2} \log |x + 1/2 + \sqrt{x^2 + x}| + c$.Note: The provided options appear to be based on a different integral $\int \sqrt{4^x + 4} \, dx$ or similar exponential forms. Given the options, the question is likely intended to be $\int \sqrt{4^x + 4} \, dx$. Assuming the intended question is $\int \sqrt{4^x + 4} \, dx$, let $u = 2^x$, then $du = 2^x \log 2 \, dx$, so $dx = \frac{du}{u \log 2}$.$I = \int \frac{\sqrt{u^2 + 4}}{u \log 2} \, du = \frac{1}{\log 2} [\sqrt{u^2 + 4} - 2 \log |\frac{2 + \sqrt{u^2 + 4}}{u}|] + c$. None of the options match the standard integral result perfectly, but option $A$ is the closest structural form for exponential integrals.

Evaluate the integral: $\int \sqrt{4x(4x + 4)} \, dx$

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