Consider the given integral.
I=∫x5x2+4dx
Let t=x2+4
dtdx=2x+0
dt2=xdx
Therefore,
I=12∫(t−4)2tdt
I=12∫t2+16−8ttdt
I=12∫(t+16t−8)dt
I=12[t22+16log(t)−8t]+C
On putting the value of t, we get
I=12⎡⎣(x2+4)22+16log(x2+4)−8(x2+4)⎤⎦+C
Hence, this is the answer.