The correct option is D x102(x5+x3+1)2+C
∫2x12+5x9(x5+x3+1)3dx
Dividing by x15in numerator and denominator
∫2x3+5x6(1+1x2+1x5)3dx Substitute1+1x2+1x5=t⇒(−2x3−5x6)dx=dt⇒(2x3+5x6)dx=−dt This gives, ∫2x3+5x6(1+1x2+1x5)3dx=∫−dtt3=12t2+C=12(1+1x2+1x5)2+C=x102(x5+x3+1)2+C