FTC #1 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}
G ( x ) = f ′ ( x ) o r ∫ a ( x ) b ( x ) d x {\displaystyle G(x)=f^{\prime }(x)or\int \limits _{a(x)}^{b(x)}\ dx}
so y = ∫ 1 − 3 x 1 1 ( 1 + u 2 ) x 3 , d x {\displaystyle y=\int \limits _{1-3x}^{1}{\frac {1}{(1+u^{2})}}x^{3},dx}
= ( 0 ) ∗ f ( 1 ) − ( − 3 ) ∗ f ( 1 − 3 x ) {\displaystyle (0)*f(1)-(-3)*f(1-3x)} ( 3 ) ∗ f ( 1 − 3 x ) {\displaystyle (3)*f(1-3x)}
= 3 ∗ ( 1 − 3 x ) 3 ∗ 1 ( 1 + ( 1 − 3 x ) 2 ) x 3 + c {\displaystyle 3*(1-3x)^{3}*{\frac {1}{(1+(1-3x)^{2})}}x^{3}+c}
