Question

How do you integrate?

Answer 1

Use the substitution `x+1=2sin2theta`.

Explanation:

Let

`I=intdx/(xsqrt(3-2x-x^2))`

Complete the square in the square root:

`I=intdx/(xsqrt(4-(x+1)^2))`

Apply the substitution `x+1=2sin2theta`:

`I=int(4cos2thetad theta)/((2sin2theta-1)(2cos2theta))`

Simplify:

`I=2int(d theta)/(4sinthetacostheta-1)`

Rearrange:

`I=2int(sec^2thetad theta)/(4tantheta-sec^2theta)`

Apply the identity `sec^2theta=tan^2theta+1`:

`I=2int(sec^2thetad theta)/(4tantheta-tan^2theta-1)`

Complete the square in the denominator:

`I=2int(sec^2thetad theta)/(3-(tantheta-2)^2)`

Apply the difference of squares `(a^2-b^2)=(a+b)(a-b)`:

`I=2int(sec^2thetad theta)/((sqrt3+tantheta-2)(sqrt3-tantheta+2))`

Apply partial fraction decomposition:

`I=1/sqrt3int(1/(sqrt3+tantheta-2)+1/(sqrt3-tantheta+2))sec^2thetad theta`

Integrate directly:

`I=1/sqrt3{ln|sqrt3+tantheta-2|-ln|sqrt3-tantheta+2|}+C`

Combine terms:

`I=1/sqrt3ln|(sqrt3+tantheta-2)/(sqrt3-tantheta+2)|+C`

Rearrange:

`I=1/sqrt3ln|(sqrt3+tantheta-2)/(sqrt3-tantheta+2)*(sqrt3-tantheta+2)/(sqrt3-tantheta+2)|+C`

Simplify:

`I=1/sqrt3ln|(3-(tantheta-2)^2)/(sqrt3-tantheta+2)^2|+C`

Rearrange:

`I=1/sqrt3ln|(3-(tantheta-2)^2)/(sqrt3-tantheta+2)^2*cos^2theta/cos^2theta|+C`

Simplify:

`I=1/sqrt3ln|(2sin2theta-1)/((2+sqrt3)costheta-sintheta)^2|+C`

Separate terms:

`I=1/sqrt3ln|2sin2theta-1|-1/sqrt3ln|((2+sqrt3)costheta-sintheta)^2|+C`

Expand:

`I=1/sqrt3ln|2sin2theta-1|-1/sqrt3ln|(2+sqrt3)^2cos^2theta+sin^2theta-2(2+sqrt3)^2sinthetacostheta|+C`

Rescale `C`:

`I=1/sqrt3ln|2sin2theta-1|-1/sqrt3ln|((2+sqrt3)^2cos^2theta+sin^2theta-2(2+sqrt3)sinthetacostheta)*(4-2sqrt3)|+C`

Simplify:

`I=1/sqrt3ln|2sin2theta-1|-1/sqrt3ln|4+2sqrt3(cos^2theta-sin^2theta)-4sinthetacostheta|+C`

Apply the double-angle trigonometric identities:

`I=1/sqrt3ln|2sin2theta-1|-1/sqrt3ln|4+2sqrt3cos2theta-2sin2theta|+C`

Reverse the substitution:

`I=1/sqrt3ln|x|-1/sqrt3ln|sqrt(9-6x-3x^2)-x+3|+C`