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

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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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