What is the integral of dx9x^2 + 1 ? a) 13 ^ 1(3x) + C b) 127 ^ 1(3x) + C c) 127 ^ 1 ( x3 ) + C d) 127 ^ 1 ( 3x ) + C

To evaluate the integral $\int \frac{dx}{9x^2+1}$, we use the standard integral formula for $\int \frac{du}{a^2+u^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{u}{a}\right)+C$.

1. Rewrite the denominator: $$9x^2+1=(3x{)}^2+1^2$$ Let $u=3x$, then $du=3dx$ and $dx=\frac{du}{3}$.

2. Substitute into the integral: $$\int \frac{dx}{9x^2+1}=\int \frac{1}{u^2+1}\cdot \frac{du}{3}=\frac{1}{3}\int \frac{du}{u^2+1}$$

3. Apply the standard integral formula: $$\frac{1}{3}\int \frac{du}{u^2+1}=\frac{1}{3}{\tan }^{-1}(u)+C=\frac{1}{3}{\tan }^{-1}(3x)+C$$

However, the correct substitution and simplification should be:

Rewrite the integral as: $$\int \frac{dx}{9x^2+1}=\frac{1}{9}\int \frac{dx}{x^2+\frac{1}{9}}$$

Let $u=3x$, then $du=3dx$ and $dx=\frac{du}{3}$: $$\frac{1}{9}\int \frac{dx}{x^2+\frac{1}{9}}=\frac{1}{9}\int \frac{1}{{\left(\frac{u}{3}\right)}^2+\frac{1}{9}}\cdot \frac{du}{3}=\frac{1}{27}\int \frac{du}{u^2+1}$$

Now apply the standard integral formula: $$\frac{1}{27}{\tan }^{-1}(u)+C=\frac{1}{27}{\tan }^{-1}(3x)+C$$

Thus, the correct answer is:

b) $\frac{1}{27}{\tan }^{-1}(3x)+C$