How to solve $\int_{-1}^{1} (x^{4/3} + 4x^{1/3}) dx$? I started by integrating it this way:
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
What is wrong with it?

The answer is

 $\frac{6}{7}$

Integrating,

$ \int^1_{-1} x^{4/3} 4 x^{1/3} \,dx = (\frac{3}{7}x^{7/3} + 3 x^{4/3})^{1}_{-1} = \frac{3}{7} + 3 + \frac{3}{7} - 3 = \frac{6}{7}. $

In the OP as it reads now, it appears that you increased the power on the second term and multiplied by the new power instead of dividing.
