Integral X per akar X+1 = …

misalkan x + 1 = u
dx = du
x = u - 1
ganti x kedalam u
= {(u - 1) / √u } du
= {(u - 1) / u^½} du
= {(u - 1)u^-½} du
= {u^(1-½) - u^-½} du
= {u^½ - u^-½} du
= (1/(½+1))u^(1+½) - (1/(-½+1))u^(1-½) + C
= (1/(3/2))u^3/2 + (1/(½))u^½ + C
= (⅔)u√u + 2√u + C
= (⅔√u)(u + 3) + C
kembalikan u = x + 1
= ⅔{√(x+1)}(x + 1 + 3) + C
= ⅔{√(x+1)}(x + 4) + C