int (x+1) / sqrt(3x^2+6x) dx Find the indefinite integral

int (x + 1)/sqrt(3x^2+6x) dx
To solve, apply u-substitution method.

u = 3x^2+6x
du = (6x+6)dx
du = 6(x + 1)dx
1/6du = (x +1)dx

Expressing the integral in terms of u, it becomes
= int 1/sqrt(3x^2 + 6x)*(x + 1)dx
= int 1/sqrtu *1/6 du
= 1/6 int1/sqrtu du
Then, convert the radical to exponent form.
= 1/6 int 1/u^(1/2)du
Also, apply the negative exponent rule a^(-m) = 1/a^m .
= 1/6 int u^(-1/2) du
To take the integral of this, apply the formula int x^n dx = x^(n+1)/(n+1)+C .
= 1/6 *u^(1/2)/(1/2) + C
= 1/6 * (2u^(1/2))/1+C
=u^(1/2)/3+C
= sqrtu /3 + C
And, substitute back u = 3x^2+6x .
= sqrt(3x^2+6x) /3 + C
 
Therefore, int (x+1)/sqrt(3x^2+6x)dx = sqrt(3x^2+6x) /3 + C .