If ∫dxx22(x7−6)=1K{ln|t|−t33−3t+3t22}+C, then which of the following is/are true
(where K is fixed constant and C is constant of integration)
A
K1008=18
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B
K1008=9
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C
t=1−6x7
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D
t=1+6x7
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Solution
The correct option is Ct=1−6x7 Let I=∫dxx22(x7−6) ⇒I=∫dxx29(1−6x−7)
Put 1−6x−7=t ⇒42x−8dx=dt ⇒I=∫1x21dxx8(1−6x−7) ⇒I=∫(1−t)363dt42t ⇒I=19072∫1−3t+3t2−t3tdt ⇒I=19072{ln|t|−3t+3t22−t33}+C ∴K1008=9 and t=1−6x7