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3.2(B) · Q78

Q.Evaluate: ∫1x2+8x+12 dx\int \frac{1}{x^2+8x+12}\,dx

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Complete the square: x2+8x+12=(x+4)2−16+12=(x+4)2−4x^2+8x+12=(x+4)^2-16+12=(x+4)^2-4.

So ∫dx(x+4)2−22\displaystyle\int\dfrac{dx}{(x+4)^2-2^2}, standard form ∫duu2−a2=12alog⁡∣u−au+a∣+c\int\dfrac{du}{u^2-a^2}=\dfrac{1}{2a}\log\left|\dfrac{u-a}{u+a}\right|+c with u=x+4, a=2u=x+4,\ a=2: …

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