Substitute x=12sect with t∈(0,π2), dx=12secttantdt:
∫√20x2−5dx=12∫√20(sect2)2−5secttantdt
∫√20x2−5dx=12∫√5sec2t−5secttantdt
∫√20x2−5dx=√52∫√sec2t−1secttantdt
Use now the trigonometric identity:
sec2t−1=tant2t
and as tanx>0 for t∈(0,π2):
√sec2t−1=tant
Then:
∫√20x2−5dx=√52∫secttan2tdt
and using the same identity:
∫√20x2−5dx=√52∫sect(sec2t−1)dt
∫√20x2−5dx=√52∫sec3tdt−√52∫sectdt
Solve now the integrals:
∫sect=ln|sect+tant|+C
and:
∫sec3tdt=∫sectddt(tant)dt
Integrate by parts:
∫sec3tdt=secttant−∫tantddt(sect)dt
∫sec3tdt=secttant−∫secttan2tdt
∫sec3tdt=secttant−∫sect(sec2t−1)dt
∫sec3tdt=secttant−∫sec3tdt+∫sectdt
Solve now for the original integral:
2∫sec3tdt=secttant+∫sectdt
∫sec3tdt=secttant2+12∫sectdt
∫sec3tdt=secttant2+12ln|sect+tant|+C
Then:
∫√20x2−5dx=√54secttant−√54ln|sect+tant|+C
Undo the substitution:
tant=√sec2t−1=√4x2−1
sect=2x
so:
∫√20x2−5dx=√52x√4x2−1−√54ln(2|x|+√4x2−1)+C
