Let I=∫x2+8x+12dx=∫x2+8x+16−4dx=∫(x+4)2−22dx(putting x+4=t⇒dx=dt)=∫t2−22dx=12tt2−22−12⋅22log⁡t+t2−22+c=12(x+4)(x+4)2−4−2log⁡x+4+(x+4)2−4+c=12(x+4)x2+8x+12−2log⁡x+4+x2+8x+12+c.

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$\int \sqrt{x^{2} + 8 x + 12} d x$

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$\frac{1}{2} (x + 4) \sqrt{x^{2} + 8 x + 12} + 2 \log |x + 4 + \sqrt{x^{2} + 8 x + 12}| + c$

$\frac{1}{2} (x + 4) \sqrt{x^{2} + 8 x + 12} - 2 \log |x + 4 + \sqrt{x^{2} + 8 x + 12}| + c$

$(x + 4) \sqrt{x^{2} + 8 x + 12} + \log |x + 4 + \sqrt{x^{2} + 8 x + 12}| + c$

$(x + 4) \sqrt{x^{2} + 8 x + 12} - \log |x + 4 + \sqrt{x^{2} + 8 x + 12}| + c$

answer is B.

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Detailed Solution

Let $I = \int \sqrt{x^{2} + 8 x + 12} d x$

$\begin{array}{l} = \int \sqrt{(x^{2} + 8 x + 16) - 4} d x \\ = \int \sqrt{(x + 4)^{2} - 2^{2}} d x \end{array}$

(putting $x + 4 = t \Rightarrow d x = d t$ )

$= \int \sqrt{t^{2} - 2^{2}} d x$

$\begin{array}{l} = \frac{1}{2} t \sqrt{t^{2} - 2^{2}} - \frac{1}{2} \cdot 2^{2} \log |t + \sqrt{t^{2} - 2^{2}}| + c \\ = \frac{1}{2} (x + 4) \sqrt{(x + 4)^{2} - 4} - 2 \log |x + 4 + \sqrt{(x + 4)^{2} - 4}| + c \end{array}$