We have,
∫5x2+4n−1√n2−4n+5dx
⇒∫5x2√n2−4n+5dx+∫4n−1√n2−4n+5dx
⇒5√n2−4n+5∫x2dx+4n−1√n2−4n+5∫1dx
⇒5x33√n2−4n+5+4n−1√n2−4n+5x+C
Hence, this is the answer.