Evaluate 
dx
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. We have,
I = 
dx
= 
dx
We find that the LCM of 2, 3 and 4 is 12. So, we substitute x = t 12 and dx = 12t 11 dt.
∴ I = 
dt
= 12 
dt
= 12 
t 5 – t 3 –2t 2 + t + 4 + 
dt
= 12 
+ 12 
dt
= (2t 6 –3t 4 –8t 3 + 6t 2 + 48t) + 12 I 1 ...(i)
where I 1 = 
dt
We have, t 3 + t + 2 = (t 3 + 1) + (t + 1) = (t + 1) (t 2 – t + 2).
So, let
= 
+ 
⇒ 3t 2 –6t + 8 = A(t 2 – t + 2) + (Bt + C) (t + 1)
Putting t = –1, 0 and 1 successively, we get
17 = 4A, 8 = 2A + C and 5 = 2A + 2B + 2C
⇒ A = 
, C = – 
and B = – 
∴ 
= 
. 
– 
⇒ I 1 = 
dt
= 
dt – 
dt
= 
log |t + 1| – 
dt
= 
log |t+1| – 
dt
= 
log |t+1| – 
– 
dt
= 
log |t+1| – 
log (t 2 – t + 2)
– 
dt
= 
log |t+1| – 
log (t 2 – t + 2) – 
tan –1 
+ C
Substituting the value of I 1 in (i), we get
I = 2t 6 –3t 4 –8t 3 + 6t 2 + 48t + 51 log |t + 1| – 15 log (t 2 –+ 2) + 
tan –1 
+ C, where t = x 1/12
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