Question

 (3x+1)dx / ((x+1)^12(3x-1)^12)

Answer 1

$\large\frac{(3x+1)dx}{((x+1)^{12}(3x-1)^{12})}$ Let's solve the problem f(x)dx is meaning of integral

$\int\large\frac{(3x+1)dx}{((x+1)^{12}(3x-1)^{12})}$

Substitute $u = (x+1)(3x-1) \to \large\frac{du}{dx}$$=3(x+1)+(3x-1) \to dx = \large\frac{du}{3(x+1)+3x-1}$

$\to dx= \large\frac{du}{2(3x+1)}$

$\large\frac{1}{2}\int\frac{du}{u^{12}}$ $\to(Apply \space power \space rule \space \int u^ndu = \large\frac{u^{n+1}}{n+1})$  with n = -12

                $= \large\frac{1}{2} * (\frac{-1}{11u^{11}})$ $=\large\frac{-1}{22u^{11}}$

Undo substitution $u = (x+1)(3x-1)$ $\to \large\frac{-1}{22(x+1)^{11}(3x-1)^{11}}$

$\large\int\large\frac{3x+1}{(x+1)^{12}(3x-1)^{12}}dx = -\large\frac{1}{22(x+1)^{11}(3x-1)^{11}}+C$