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

The integral of 1/((3x)^2)+3 is -1/(9x)+3x+C

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

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

- \frac{1}{9 x} + 3 x + C

Solution steps

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

Simplify

\frac{1}{( 3 x ) ^{2}} + 3 : \frac{1}{9 x ^{2}} + 3

= \int \frac{1}{9 x ^{2}} + 3 d x

Apply the Sum Rule:

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

= \int \frac{1}{9 x ^{2}} d x + \int 3 d x

\int \frac{1}{9 x ^{2}} d x = - \frac{1}{9 x}

\int 3 d x = 3 x

= - \frac{1}{9 x} + 3 x

Add a constant to the solution

= - \frac{1}{9 x} + 3 x + C

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What is the integral of 1/((3x)^2)+3 ?
The integral of 1/((3x)^2)+3 is -1/(9x)+3x+C