`int 1/sqrt(x^2-4) dx` Find the indefinite integral

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`int1/sqrt(x^2-4)dx`

Let's apply integral substitution:`x=2sec(u)`

`=>dx=2sec(u)tan(u)du`

`=int1/sqrt((2sec(u))^2-4)2sec(u)tan(u)du`

`=int(2sec(u)tan(u))/sqrt(4sec^2(u)-4)du`

`=int(2secutan(u))/(sqrt(4)sqrt(sec^2(u)-1))du`

Now use the trigonometric identity: `tan^2(x)=sec^2(x)-1`

`=int(2sec(u)tan(u))/(2sqrt(tan^2(u)))du`

`=intsec(u)du`

Now use the standard integral:`intsec(x)dx=ln|sec(x)+tan(x)|`

`=ln|sec(u)+tan(u)|` ----------(1)

Now from the substitution:

`sec(u)=x/2`  and,

`tan^2(u)=sec^2(u)-1`

`tan^2(u)=(x/2)^2-1`

`tan^2(u)=(x^2-4)/4`

`tan(u)=sqrt(x^2-4)/2`

Substitute back the above in the result  (1)

`=ln|x/2+sqrt(x^2-4)/2|`

`=ln|(x+sqrt(x^2-4))/2|`

`=ln|x+sqrt(x^2-4)|-ln(2)`

Since `ln(2)`  is constant, so we can...

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`int1/sqrt(x^2-4)dx`

Let's apply integral substitution:`x=2sec(u)`

`=>dx=2sec(u)tan(u)du`

`=int1/sqrt((2sec(u))^2-4)2sec(u)tan(u)du`

`=int(2sec(u)tan(u))/sqrt(4sec^2(u)-4)du`

`=int(2secutan(u))/(sqrt(4)sqrt(sec^2(u)-1))du`

Now use the trigonometric identity: `tan^2(x)=sec^2(x)-1`

`=int(2sec(u)tan(u))/(2sqrt(tan^2(u)))du`

`=intsec(u)du`

Now use the standard integral:`intsec(x)dx=ln|sec(x)+tan(x)|`

`=ln|sec(u)+tan(u)|` ----------(1)

Now from the substitution:

`sec(u)=x/2`  and,

`tan^2(u)=sec^2(u)-1`

`tan^2(u)=(x/2)^2-1`

`tan^2(u)=(x^2-4)/4`

`tan(u)=sqrt(x^2-4)/2`

Substitute back the above in the result  (1)

`=ln|x/2+sqrt(x^2-4)/2|`

`=ln|(x+sqrt(x^2-4))/2|`

`=ln|x+sqrt(x^2-4)|-ln(2)`

Since `ln(2)`  is constant, so we can omit it and add a new constant C to the solution,

`=ln|x+sqrt(x^2-4)|+C`

 

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