Integrate $\int \frac{x^3}{3+3x^2} $

Question

Integrate $$\int \frac{x^3}{3+3x^2} $$

I want to solve it using only this formula-

$$\int \frac{ f(x)}{f’(x)} dx = \ln (f(x)) $$

$f(x) = 3+3x^2$

$f’(x) = 6x$

Therefore, $\frac{1}{6} (x^2) \int \frac{6x}{3+ 3x^2} dx = \frac{1}{6} (x^2) \ln (3+3x^2) + C $

why is my answer wrong ?

Answer 1

But $$\ln(f(x))$$ has the first derivative $$\frac{f'(x)}{f(x)}$$

Answer 2

Your formula:

$$\int \frac{ f(x)}{f’(x)} dx = \ln (f(x))$$

...is wrong! It should be:

$$\int \frac{ f'(x)}{f(x)} dx = \ln (f(x))$$

Answer 3

\begin{align} \int{\frac{x^3}{3+3x^2}\ dx} &=\frac13\int{\left(x-\frac{x}{x^2+1}\right)\ dx}\\ &=\frac13\int{\left(x-\frac12\frac{2x}{x^2+1}\right)\ dx}\\ &\quad\vdots\\ &=\frac16\left(x^2-\ln(x^2+1)\right)+C \end{align}