How do you integrate #int (3x+4)/(sqrt(3x^2+8x+3))# using substitution?

To integrate ( \int \frac{3x+4}{\sqrt{3x^2+8x+3}} ) using substitution, let ( u = \sqrt{3x^2+8x+3} ). Then, ( u^2 = 3x^2 + 8x + 3 ).

Differentiate both sides of ( u = \sqrt{3x^2+8x+3} ) with respect to ( x ) to find ( du ):

( \frac{du}{dx} = \frac{1}{2\sqrt{3x^2+8x+3}} \cdot (6x + 8) )

Rearrange to solve for ( dx ): ( dx = \frac{2du}{6x + 8} )

Substitute ( u = \sqrt{3x^2+8x+3} ) and ( dx = \frac{2du}{6x + 8} ) into the integral:

( \int \frac{3x+4}{u} \cdot \frac{2du}{6x + 8} )

Simplify:

( \int \frac{6x+8}{3u} \cdot \frac{du}{6x + 8} )

( \int \frac{1}{u} , du )

This is a straightforward integral of ( \frac{1}{u} ), which is ( \ln|u| + C ).

Substitute back for ( u ):

( \ln|\sqrt{3x^2+8x+3}| + C )

Thus, ( \int \frac{3x+4}{\sqrt{3x^2+8x+3}} , dx = \ln|\sqrt{3x^2+8x+3}| + C ), where ( C ) is the constant of integration.