Evaluate the integral: int sqrt{x^2+4x+1} , d | vedclass

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Question

(A) To evaluate $\int \sqrt{x^2+4x+1} \, dx$,first complete the square inside the square root:$x^2+4x+1 = (x^2+4x+4) - 3 = (x+2)^2 - (\sqrt{3})^2$.Now

Vedclass · Accepted Answer

$\frac{x+2}{2} \sqrt{x^2+4x+1} - \frac{3}{2} \log \left|x+2+\sqrt{x^2+4x+1}\right|$

Vedclass · Answer

(A) To evaluate $\int \sqrt{x^2+4x+1} \, dx$,first complete the square inside the square root:$x^2+4x+1 = (x^2+4x+4) - 3 = (x+2)^2 - (\sqrt{3})^2$.Now the integral becomes $\int \sqrt{(x+2)^2 - (\sqrt{3})^2} \, dx$.Using the standard formula $\int \sqrt{t^2-a^2} \, dt = \frac{t}{2} \sqrt{t^2-a^2} - \frac{a^2}{2} \log |t + \sqrt{t^2-a^2}| + C$,where $t = x+2$ and $a = \sqrt{3}$:$= \frac{x+2}{2} \sqrt{(x+2)^2 - 3} - \frac{3}{2} \log |(x+2) + \sqrt{(x+2)^2 - 3}| + C$.$= \frac{x+2}{2} \sqrt{x^2+4x+1} - \frac{3}{2} \log |x+2 + \sqrt{x^2+4x+1}| + C$.Comparing this with the given options,the correct option is $A$.

Evaluate the integral: $\int \sqrt{x^2+4x+1} \, dx = \text{ . . . . . . } + C$.

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