Respuesta :

Nirina7
let be y= world, and joy=x
so we have integral (world)joy^world-1d(joy) = integ(yx^y-1)dx, y is a constant independant of x, so integ(yx^y-1)dx = y . integ(x^y-1)dx
=y .(x^y-1+1 / y-1 +1) + C=(x^y) + C

the answer is joy^world

Answer:

The answer is [tex]joy^{world} + C[/tex]

Step-by-step explanation:

Given the indefinite integral [tex]\int (world)joy^{world-1}d(joy)[/tex]

Let world → a

         joy → x

Therefore, the integral becomes [tex]\int ax^{a-1}dx[/tex]

                                                    = [tex]a\frac{x^{a-1+1} }{a-1+1} + C[/tex]

                                                    = [tex]x^{a} + C[/tex]

Replacing, a → world

                  x → joy  

Hence, [tex]\int (world)joy^{world-1}d(joy)[/tex] = [tex]joy^{world} + C[/tex]

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