`int (x^2+3)/(xsqrt(x^2-4)) dx` Find the indefinite integral

Recall indefinite integral follows `int f(x) dx = F(x)+C`

 where:

`f(x)` as the integrand

`F(x)` as the antiderivative of `f(x)`

`C` as the constant of integration.

 The given problem: `int (x^2+3)/(xsqrt(x^2-4)) dx` has an integrand of `f(x)=(x^2+3)/(xsqrt(x^2-4))` .

Apply u-substitution on `f(x) dx` by letting `u =x^2` then `du = 2x dx` or `dx= (du)/(2x)` :

`int (x^2+3)/(xsqrt(x^2-4))dx =int (u+3)/(xsqrt(u-4))*(du)/(2x)`

                        `=int ((u+3)du)/(2x^2sqrt(u-4))`

                         `=int ((u+3)du)/(2usqrt(u-4))`

Apply the basic integration property:` int c*f(x) dx = c int f(x) dx` :

`int ((u+3)du)/(2usqrt(u-4))=(1/2)int ((u+3)du)/(usqrt(u-4))`

Apply the basic integration property for sum:

`int (u+v) dx = int (u) dx+int (v) dx.`

`(1/2)int ((u+3)du)/(usqrt(u-4))=(1/2) [int (udu)/(usqrt(u-4))+int (3du)/(usqrt(u-4))]`

For the integration of the`int (udu)/(usqrt(u-4))` , we can cancel out the u:

`int (udu)/(usqrt(u-4))=int (du)/sqrt(u-4)`

Let ` v= u-4` then `dv =du` .

Apply the Law of exponents: `sqrt(x)= x^1/2` and `1/x^n= x^-n` ,  we get:

`int (du)/sqrt(u-4)=int (dv)/sqrt(v)`

Apply the Power Rule for integration: `int x^n dx= x^(n+1)/(n+1)+C`

`int v^(-1/2)dv=v^((-1)/2+1)/((-1)/2+1) +C`

                  ` = v^(1/2)/(1/2)`

                 ` =v^(1/2)*(2/1)`

                 ` = 2v^(1/2)`  or   `2sqrt(v)`

With `v= u-4`  then `2sqrt(v) = 2sqrt(u-4)` .

The integral becomes:

`int (du)/sqrt(u-4)=2sqrt(u-4).`

For the integration of `int (3du)/(usqrt(u-4))` , we basic integration property: `int c*f(x) dx = c int f(x)`

`int (3du)/(usqrt(u-4))=3int (du)/(usqrt(u-4))`

Let: `v= sqrt(u-4)`

Then square both sides to get `v^2=u-4` then `v^2+4 =u`

Applying implicit differentiation on `v^2=u-4` , we get: `2vdv = du` .

Plug-in `du =2v dv` , ` u=v^2+4` and `v=sqrt(u-4)` , we get:

`3 int (du)/(usqrt(u-4))=3int (2vdv)/((v^2+4)*v)`

                   ` =3int (2dv)/((v^2+4))`

                   ` =3*2int (dv)/(v^2+4)`

                   ` =6int (dv)/(v^2+4)`

The integral part resembles the basic integration for inverse tangent function:

`int (dx)/(x^2+a^2) = (1/a)arctan(u/a)+C`

Then,

`6int (dv)/(v^2+4) =6*(1/2)arctan(v/2)+C`

                 `=3arctan(v/2)+C`

Plug-in `v =sqrt(u-4)` , we get:

`int (3du)/(usqrt(u-4)) =3arctan(sqrt(u-4)/2)+C`

Combining the results, we get:

`(1/2) [int (udu)/(usqrt(u-4))+int (3du)/(usqrt(u-4))] = (1/2)*[2sqrt(u-4)+3arctan(sqrt(u-4)/2)]+C`

` =sqrt(u-4)+3/2arctan(sqrt(u-4)/2)+C`

Plug-in `u = x^2` to get the final answer:

`int (x^2+3)/(xsqrt(x^2-4)) dx= sqrt(x^2-4)+3/2arctan(sqrt(x^2-4)/2)+C`