Integral 2x akar x - 1 dx

jawab

∫ 2x √(x-1) dx
u =x - 1 → x = u + 1
2x = 2(u+1)

u = x  - 1
du /dx = 1
dx = du
[tex] \int\limits {2x} \sqrt{x-1} \,dx \ = \int\limits {2(u+1) u^{ \frac{1}{2} } } \, dx \\ \ = 2\int\limits {u^{ \frac{3}{2} }+ u^{ \frac{1}{2} } } \, dx \\ \ = 2\ ( \frac{2}{5}\ u^{ \frac{5}{2} } + \frac{2}{3} \ u^{ \frac{3}{2} } )\\ \,= \frac{4}{5}\ u^{ \frac{5}{2} }+ \frac{4}{3} u^{ \frac{3}{2}} \\ \,= \frac{4}{15}\, u^{ \frac{3}{2} } (u + 1) + c\\ \,= \frac{4}{15}(u +1) (u \sqrt{u} ) + c\\ \, = \frac{4}{15}(x)(x-1) \sqrt {x -1} + c [/tex]