Question

How do you integrate?

`int 1/(x+sqrt(x^2-2x-3))dx`

Answer 1

`int1/(x+sqrt(x^2-2x-3))dx=1/2(x-sqrt(x^2-2x-3))+5/4ln|x-1+sqrt(x^2-2x-3)|-3/4ln|(2x+3)*((3x+3+sqrt(x^2-2x-3))/(3x+3-sqrt(x^2-2x-3)))|+C`

Explanation:

Let

`I=int1/(x+sqrt(x^2-2x-3))dx`

Rationalize:

`I=int1/(x+sqrt(x^2-2x-3))*(x-sqrt(x^2-2x-3))/(x-sqrt(x^2-2x-3))dx`

Simplify:

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

Rearrange:

`I=1/2int(1-3/(2x+3))dx-intsqrt(x^2-2x-3)/(2x+3)dx`

Complete the square in the square root:

`I=1/2(x-3/2ln|2x+3|)-intsqrt((x-1)^2-4)/(2x+3)dx`

Apply the substitution `x-1=2sectheta`:

`I=1/2x-3/4ln|2x+3|-int(2tantheta)/(4sectheta+5)(2secthetatanthetad theta)`

Simplify:

`I=1/2x-3/4ln|2x+3|-4int(tan^2theta/(4sectheta+5))secthetad theta`

Since `tan^2theta=sec^2theta-1`:

`I=1/2x-3/4ln|2x+3|-1/4int(4sectheta-5+9/(4sectheta+5))secthetad theta`

Rearrange:

`I=1/2x-3/4ln|2x+3|-1/4int(4sec^2theta-5sectheta)d theta-9/4int1/(4+5costheta)d theta`

Apply the trigonometric identity `costheta=2cos^2(theta/2)-1`:

`I=1/2x-3/4ln|2x+3|-1/4(4tantheta-5ln|2sectheta+2tantheta|)-9/4int1/(10cos^2(theta/2)-1)d theta`

Rearrange:

`I=1/2x-3/4ln|2x+3|-1/2(2tantheta)+5/4ln|2sectheta+2tantheta|-9/4intsec^2(theta/2)/(9-tan^2(theta/2))d theta`

Apply the difference of squares and partial fraction decomposition:

`I=1/2x-3/4ln|2x+3|-1/2(2tantheta)+5/4ln|2sectheta+2tantheta|-3/8int(1/(3+tan(theta/2))+1/(3-tan(theta/2)))sec^2(theta/2)d theta`

Integrate directly:

`I=1/2x-3/4ln|2x+3|-1/2(2tantheta)+5/4ln|2sectheta+2tantheta|-3/4(ln|3+tan(theta/2)|-ln|3-tan(theta/2)|)+C`

Since `tan(theta/2)=tantheta/(1+sectheta)`:

`I=1/2x-3/4ln|2x+3|-1/2(2tantheta)+5/4ln|2sectheta+2tantheta|-3/4ln|(6+3(2sectheta)+2tantheta)/(6+3(2sectheta)-2tantheta)|+C`

Reverse the substitution:

`I=1/2x-3/4ln|2x+3|-1/2sqrt(x^2-2x-3)+5/4ln|x-1+sqrt(x^2-2x-3)|-3/4ln|(3x+3+sqrt(x^2-2x-3))/(3x+3-sqrt(x^2-2x-3))|+C`

Collect terms:

`I=1/2(x-sqrt(x^2-2x-3))+5/4ln|x-1+sqrt(x^2-2x-3)|-3/4ln|(2x+3)*((3x+3+sqrt(x^2-2x-3))/(3x+3-sqrt(x^2-2x-3)))|+C`