What is the integral of (x+1)/(sqrt(3x^2+6x)) ?

The integral of (x+1)/(sqrt(3x^2+6x)) is 1/3 sqrt(3x^2+6x)+C

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integral of (x+1)/(sqrt(3x^2+6x))

\int \frac{x + 1}{3 x ^{2} + 6 x} d x

\frac{1}{3} 3 x^{2} + 6 x + C

Solution steps

\int \frac{x + 1}{3 x ^{2} + 6 x} d x

Apply u-substitution

= \int \frac{1}{6 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Apply the Power Rule

= \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 3 x^{2} + 6 x

= \frac{1}{6} \cdot 2 (3 x^{2} + 6 x)^{\frac{1}{2}}

Simplify

\frac{1}{6} \cdot 2 (3 x^{2} + 6 x)^{\frac{1}{2}} : \frac{1}{3} 3 x^{2} + 6 x

= \frac{1}{3} 3 x^{2} + 6 x

Add a constant to the solution

= \frac{1}{3} 3 x^{2} + 6 x + C

What is the integral of (x+1)/(sqrt(3x^2+6x)) ?
The integral of (x+1)/(sqrt(3x^2+6x)) is 1/3 sqrt(3x^2+6x)+C