How do you find the integral of √x2−25x?

1 Answer
Andrea S.
Mar 31, 2018

∫√x2−25xdx=√x2−25−arctan(√x2−255)+C

Explanation:

The function is defined for |x|≥5. Restrict for the moment to x>5 and substitute:

x=5sect

dx=5secttantdt

with t∈[0,π2).

Then:

∫√x2−25xdx=5∫√25sec2t−25secttant5sectdt

∫√x2−25xdx=5∫√sec2t−1tantdt

Use now the trigonometric identity:

sec2−1=tan2t

and as for ∈[0,π2) the tangent is positive:

√sec2−1=tant

so:

∫√x2−25xdx=5∫tan2tdt

using the same identity again:

∫√x2−25xdx=5∫(sec2t−1)dt

and for the linearity of the integral:

∫√x2−25xdx=5∫sec2tdt−5∫dt

∫√x2−25xdx=5tant−5t+C

To undo the substitution note that:

tant=√sec2t−1=√(x5)2−1=15√x2−25

and then:

t=arctan(√x2−255)

So:

∫√x2−25xdx=√x2−25−arctan(√x2−255)+C

By differentiating both sides we can verify that the solution extends also to x<−5.

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