# 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

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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 .

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