∫dxx2x4+13/4=Ax4+1x4B+C
A=−1,B=14
A=1,B=−14
A=12,B=1
A=−12,B=−1
I=∫dxx2x4+13/4=∫dxx2⋅x31+1x43/4
Put1+x−4=t⇒−4x5dx=dt⇒−14∫dtt3/4=14∫t−3/4dt=−14⋅t1/41/4+C=−1+1x41/4+C=−1+x4x41/4+C
Hence, A=−1,B=14