Answer:
Convert to the form of fundamental theorem
[tex]y = \int_{1-6x}^1 \frac{u^3}{1+u^2}du= -\int\limits^{1-6x}_1 {\frac{u^3}{1+u^2} } \, du[/tex]
Step-by-step explanation:
Now, you can apply the theorem, that is [tex]y'= f(1-6x) \times (1-6x)' = -6 \frac{(1-6x)^3}{(1+(1-6x))}[/tex]
Simplify!!
