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

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

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

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

x (x + 1) + C

Solution steps

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

Take the constant out:

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

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

Expand

\frac{6 x ^{2} + 3 x}{x} : \frac{6 x ^{2}}{x} + \frac{3 x}{x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{6 x ^{2}}{x} d x = 3 x^{2}

\int \frac{3 x}{x} d x = 3 x

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

Simplify

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

= x (x + 1)

Add a constant to the solution

= x (x + 1) + C

\int (3 x^{2})^{2} d x

\int 2 sec^{2} (θ) tan (θ) d θ

\int \frac{x ^{3}}{3 x ^{4} - 5} d x

\int \frac{2 x ^{3} + 9 x ^{2} + 14 x + 8}{( x ^{2} + 2 x ) ( x ^{2} + 2 )} d x

\frac{1}{x + 1}

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