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3.2(B) · Q81

Q.Evaluate: ∫15−4x−3x2 dx\int \frac{1}{5-4x-3x^2}\,dx

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Write 5−4x−3x2=−3(x2+43x)+5=−3(x+23)2+43+5=193−3(x+23)2=3[199−(x+23)2]5-4x-3x^2=-3\left(x^2+\dfrac43x\right)+5=-3\left(x+\dfrac23\right)^2+\dfrac43+5=\dfrac{19}3-3\left(x+\dfrac23\right)^2=3\left[\dfrac{19}9-\left(x+\dfrac23\right)^2\right].

So ∫dx5−4x−3x2=13∫dua2−u2\displaystyle\int\dfrac{dx}{5-4x-3x^2}=\dfrac13\int\dfrac{du}{a^2-u^2} with u=x+23, a=193u=x+\dfrac23,\ a=\dfrac{\sqrt{19}}3: …

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