5.3 The Fundamental Theorem of Calculus/17

FTC #1 $y=\int \limits _{1-3x}^{1}{\frac {x^{3}}{(1+u^{2})}}dx$

$G(x)=f^{\prime }(x)$ or Failed to parse (syntax error): {\displaystyle \(d/dx)(int\limits_{a(x)}^{b(x)})\ b(x)*f(b(x))-a(x)*f(a(x) dx}

so $y=\int \limits _{1-3x}^{1}{\frac {1}{(1+u^{2})}}x^{3},dx$

=$(0)*f(1)-(-3)*f(1-3x)$ $(3)*f(1-3x)$

=$3*(1-3x)^{3}*{\frac {1}{(1+(1-3x)^{2})}}x^{3}+c$