# Use the specified substitution to find or evaluate the integral. `int sqrt(x-2)/(x+1) dx` `u = sqrt(x-2)`

lemjay | Certified Educator

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

`u = sqrt(x-2)`

To be able to apply substitution method, isolate the x in our assumption.

`u=sqrt(x-2)`

`u^2=x - 2`

`u^2+2=x`

Then, differentiate both sides.

`2u du = dx`

So, plug-in `u=sqrt(x-2)` , `x=u^2+2` and `dx = 2u du` to the integral.

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

`= int u/(u^2+2+1) (2udu)`

`=int (2u^2)/(u^2+3)du`

Then, expand the integrand

`= int(2 -6/(u^2+3)du`

`=int 2du -int 6/(u^2+3)du`

For the first integral apply the formula `int kdx = kx + C` .

For the second integral, apply the formula ` int 1/(a^2+x^2)dx = 1/a tan^(-1)(x/a)+C` .

`=int 2du - int 6(u^2+(sqrt3)^2)du`

`=2u - 6 (1/sqrt3 tan^(-1) (u/sqrt3)) + C`

`=2u - 6/sqrt3 tan^(-1) (u/sqrt3)+C`

To simplify it further, rationalize the denominator.

`=2u - 6/sqrt3 *sqrt3/sqrt3tan^(-1) (u/sqrt3*sqrt3/sqrt3)+C`

`=2u-2sqrt3tan^(-1)((usqrt3)/3)+C`

And, substitute back `u =sqrt(x+2)` .

`=2sqrt(x+2) -2sqrt3tan^(-1)( (sqrt(x+2)sqrt3)/3)+C`

`=2sqrt(x+2) -2sqrt3tan^(-1)(sqrt(3(x+2))/3)+C`

Therefore, `int sqrt(x-2)/(x+1) dx=2sqrt(x+2) -2sqrt3tan^(-1)(sqrt(3(x+2))/3)+C` .