How do you integrate?

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Answer 1 · 2018-03-11T20:31:45

Use the substitution $x+3=3sectheta$ .

Let

$I=int(x-3)/sqrt(x^2+6x)dx$

Complete the square in the square root:

$I=int(x-3)/sqrt((x+3)^2-9)dx$

Apply the substitution $x+3=3sectheta$ :

$I=int(3sectheta-6)/(3tantheta)(3secthetatanthetad theta)$

Simplify:

$I=3int(sec^2theta-2sectheta)d theta$

Integrate directly:

$I=3(tantheta-2ln|sectheta+tantheta|)+C$

Rearrange:

$I=3tantheta-6ln|3sectheta+3tantheta|+C$

Reverse the substitution:

$I=sqrt(x^2+6x)-6ln|x+3+sqrt(x^2+6x)|+C$