FTC #1
G ( x ) = f ′ ( x ) {\displaystyle G(x)=f^{\prime }(x)} or in other words d d x [ ∫ a ( x ) b ( x ) F ( x ) d x ] {\displaystyle {\frac {d}{dx}}[\int \limits _{a(x)}^{b(x)}F(x)dx]} is b ′ ( x ) ∗ f ( b ( x ) ) − a ′ ( x ) ∗ f ( a ( x ) ) {\displaystyle \ b^{\prime }(x)*f(b(x))-a^{\prime }(x)*f(a(x))}
y = ∫ 1 − 3 x 1 u 3 ( 1 + u 2 ) d u {\displaystyle y=\int \limits _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du}
so y = ∫ 1 − 3 x 1 1 ( 1 + u 2 ) u 3 , d u {\displaystyle y=\int \limits _{1-3x}^{1}{\frac {1}{(1+u^{2})}}u^{3},du}
using the formula we get y= ( 0 ) ∗ f ( 1 ) − ( − 3 ) ∗ f ( 1 − 3 x ) {\displaystyle (0)*f(1)-(-3)*f(1-3x)}
which is equal to ( 3 ) ∗ f ( 1 − 3 x ) {\displaystyle (3)*f(1-3x)}
which is= 3 ∗ ( 1 − 3 x ) 3 ∗ 1 ( 1 + ( 1 − 3 x ) 2 ) {\displaystyle 3*(1-3x)^{3}*{\frac {1}{(1+(1-3x)^{2})}}}
or simplified to 3 ∗ ( 1 − 3 x ) 3 ( 1 + ( 1 − 3 x ) 2 ) {\displaystyle {\frac {3*(1-3x)^{3}}{(1+(1-3x)^{2})}}}
FTC #2
∫ a b f ( x ) d x {\displaystyle \int \limits _{a}^{b}f(x)dx} is equal to F ( b ) − F ( a ) {\displaystyle F(b)-F(a)} Where F is the antiderivative of f such that F ′ = f {\displaystyle F^{\prime }=f}
y = ∫ 1 − 3 x 1 x 3 ( 1 + u 2 ) d x {\displaystyle y=\int \limits _{1-3x}^{1}{\frac {x^{3}}{(1+u^{2})}}dx}
or in other words ![{\displaystyle {\frac {d}{dx}}[\int \limits _{a(x)}^{b(x)}F(x)dx]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ba58f09e91a121dd850cf63fc93a0c2b6af88084)
is 
is equal to 
Where F is the antiderivative of f such that 
