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3.2(B) · Q69

Q.Evaluate: ∫14x2−3 dx\int \frac{1}{4x^2-3}\,dx

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✓ Free question

Write 4x2−3=4(x2−34)4x^2-3=4\left(x^2-\dfrac34\right), so ∫dx4x2−3=14∫dxx2−(32)2\int\dfrac{dx}{4x^2-3}=\dfrac14\int\dfrac{dx}{x^2-\left(\frac{\sqrt3}{2}\right)^2}.

This is the standard form ∫dxx2−a2=12alog⁡∣x−ax+a∣+c\int\dfrac{dx}{x^2-a^2}=\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c with a=32a=\dfrac{\sqrt3}{2}.

So the integral =14⋅12⋅32log⁡∣x−32x+32∣+c=143log⁡∣x−32x+32∣+c=\dfrac14\cdot\dfrac{1}{2\cdot\frac{\sqrt3}2}\log\left|\dfrac{x-\frac{\sqrt3}2}{x+\frac{\sqrt3}2}\right|+c=\dfrac{1}{4\sqrt3}\log\left|\dfrac{x-\frac{\sqrt3}2}{x+\frac{\sqrt3}2}\right|+c.

Multiplying numerator and denominator of the fraction inside the log by 2:

✓Final answer

143log⁡∣2x−32x+3∣+c\dfrac{1}{4\sqrt3}\log\left|\dfrac{2x-\sqrt3}{2x+\sqrt3}\right|+c

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