Question

# ∫(4x−1)dx√2x2−6x+18 is equal to

A
22x26x+1852ln((x32)+x23x+9+C
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B
22x26x+18+52ln((x32)+x23x+9+C
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C
22x26x+1852ln((x32)+x23x+9+C
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D
22x26x+18+52ln((x32)+x23x+9+C
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Solution

## The correct option is D 2√2x2−6x+18+5√2ln((x−32)+√x2−3x+9∣∣+CSolving integrals of these types recquires expressing the numerator of the integrand as the sum of derivative of the denominator and a constant. So here, we can express the numerator as: (4x−1)=addx(2x2−6x+18)+b ⇒(4x−1)=a(4x−6)+b Thus, comparing the coefficient of x and constant term we get a=1, b=5 Now ,we can write the integral as: ⇒I=∫(4x−1)dx√2x2−6x+18⇒I=∫(4x−6)+5√2x2−6x+18dx⇒I=∫(4x−6)√2x2−6x+18dx +5∫dx√2x2−6x+18⇒I=2√2x2−6x+18+I1 Now, we can evaluate I1 as: I1=5∫dx√2x2−6x+18⇒I1=5√2∫dx√x2−3x+9⇒I1=5√2∫dx√x2−3x+94−94+9⇒I1=5√2∫dx√(x−32)2−274⇒I1=5√2∫dx√(x−32)2−(√272)2Now, using the formula ∫dx√x2−a2=ln(x+√x2−a2∣∣+C,we get⇒I1=5√2ln((x−32)+√x2−3x+9∣∣ +C Thus the integral is I=2√2x2−6x+18+ 5√2ln((x−32)+√x2−3x+9∣∣ +C Thus, Option d. is correct.

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