Question

How do you integrate?

`int_sqrt3^2(sqrt(x^2-3))/xdx`

Answer 1

`int_sqrt3^2(sqrt(x^2-3))/xdx=1-(pisqrt3)/6`

Explanation:

We could use trigonometric substitution to deal with this integral.

Draw a right triangle with angle `theta`.

Label the hypotenuse as `x`.

Label the side adjacent to `theta` as `sqrt3`

Now we can write

`sectheta=x/sqrt3`

`rArrcolor(green)(x=sqrt3sectheta)`

`rArrcolor(red)(dx=sqrt3secthetatantheta` `color(red)d` `color(red)theta`

Also, by the Pythagorean Theorem, the side of the triangle opposite to `theta` is `sqrt(x^2-3)`. So we can write

`tantheta=sqrt(x^2-3)/sqrt3`

`rArrcolor(blue)(sqrt(x^2-3)=sqrt3tantheta)`

Now let's write the integral in terms of `theta`

`int_sqrt3^2color(blue)sqrt(x^2-3)/color(green)xcolor(red)dxrArrint_sqrt3^2color(blue)(sqrt3tantheta)/color(green)(sqrt3sectheta)color(red)(sqrt3secthetatantheta` `color(red)d` `color(red)theta)`

Now simplify

`rArrint_sqrt3^2(sqrt3tantheta)/cancel(sqrt3sectheta)cancel(sqrt3sectheta)tantheta` `d` `theta`

`rArrint_sqrt3^2sqrt3tan^2theta` `d` `theta`

Use the identity `sec^2theta=1+tan^2theta`

`rArrsqrt3int_sqrt3^2(sec^2theta-1)` `d` `theta`

Now we can take the integral

`rArrsqrt3[tantheta-theta]_sqrt3^2`

Put it back in terms of `x`

`rArrsqrt3[(sqrt(x^2-3))/sqrt3-arctan((sqrt(x^2-3))/sqrt3)]_sqrt3^2`

Now let's evaluate the endpoints:

`rArrsqrt3[(sqrt(4-3)/sqrt3-arctan(sqrt(4-3)/sqrt3))-(sqrt(3-3)/sqrt3-arctan(sqrt(3-3)/sqrt3))]`

`rArrsqrt3[(1/sqrt3-arctan(1/sqrt3))-(0-arctan(0))]`

`rArrsqrt3[1/sqrt3-pi/6]`

`rArr1-(pisqrt3)/6`

Answer 2

In the given problem

`I=int\ sqrt(x^2−3)/xdx`

Substitute `u=sqrt(x^2−3)`
Differentiating both sides with respect to respective variables we get

`du=x/sqrt(x^2−3)dx`
`=> dx=sqrt(x^2−3)/xdu`

`:.I=int\ (sqrt(x^2−3)/x)(sqrt(x^2−3)/xdu)`
`=>int\ u^2/(u^2+3)du`

Rewriting numerator as `(u^2+3)-3` and simplifying we get

`I=int\ (1-3/(u^2+3))du`
`I=I_1-I_2=int\ du-int3/(u^2+3)du`

Now

`I_1=u`

and by dividing both numerator and denominator with `3,` `I_2` can be rewritten as

`I_2=int\ 1/((u/sqrt3)^2+1)du`

substitute `v=u/sqrt3` to make it a standard integral. We get

`dv=(du)/sqrt3`
`:.I_2=sqrt3int\ 1/(v^2+1)dv`
`=>I_2=sqrt3tan^-1v` `" (using standard integral")`

Making substitution back to `u` we get

`I=u-sqrt3tan^-1(u/sqrt3)+C`
where `C` is a constant of integration.

Making substitution back to `x` and applying limits we get

`I=[sqrt(x^2−3)-sqrt3tan^-1((sqrt(x^2−3))/sqrt3)+C]_sqrt3^2`
`I=[sqrt(2^2−3)-sqrt3tan^-1((sqrt(2^2−3))/sqrt3)+C]-[sqrt((sqrt3)^2−3)-sqrt3tan^-1((sqrt((sqrt3)^2−3))/sqrt3)+C]`
`=>I=1-sqrt3tan^-1(1/sqrt3)`
`=>I=1-sqrt3pi/6`