∫x2-1x32x4-2x2+1dx=
-122–2x2+1x4+C
12log(2x4–2x2+1)+C
122–2x2+1x4+C
None of these
Explanation for the correct option:
Evaluating the integration:
∫x2-1x32x4-2x2+1dx
⇒∫x2-1x3x42-2x2+1x4dx[factoroutx4indenominator]⇒∫x2-1x52-2x2+1x4dx...(i)∴x4=x2
Substituting t=2-2x2+1x4...(ii)
Differentiating t w.r.t. x
⇒-4x5+4x3dx=dt⇒4x2-1x5dx=dt......(iii)
⇒14∫dtt=14∫t-12dt[from(i),(ii)&(iii)]=14t-1212+C[C=constant]=12t+C....(iv)
Substituting back t=2-2x2+1x4 in equation (iv)
=122-2x2+1x4+C
Hence, option (C) is the correct option.
For each pair of polynomials p(x) and q(x) given below find the degree of p(x) + q(x) and p(x)q(x).
(i)
(ii)
(iii)
(iv)