What is the antiderivative of #sqrt(x+3)#?

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Answer 1 · 2018-05-02T15:22:08

#2/3(x+3)^(3/2)+C#

#intsqrt(x+3)dx#

#color(green)(int(x+a)^ndx=(x+a)^(n+1)/(n+1)+C)#

#intsqrt(x+3)dx=color(blue)((x+3)^(1/2+1)/(1/2+1)+C#

#=2/3(x+3)^(3/2)+C#

Answer 2 · 2018-05-02T15:24:25

#intsqrt(x+3)dx=2/3(x+3)^(3/2)+"c"#

Finding the antiderivative of a function is the same as finding its integral (by the Fundamental Theorem of Calculus).

To find #intsqrt(x+3)dx#, we can use recognition or a natural substitution. We will use the latter.

Let #u=x+3# and #du=dx#. Then

#intsqrt(x+3)dx=intsqrtudu=intu^(1/2)du#

Now we employ the power rule for integration:

#intx^ndx=1/(n+1)x^(n+1)+"c"#

Thus

#intu^(1/2)du=2/3u^(3/2)+"c"=2/3(x+3)^(3/2)+"c"#

Answer 3 · 2018-05-02T15:26:16

The answer #2/3(x+1)^(3/2)+c#

#intsqrt(x+3)*dx=int(x+1)^(1/2)*dx=2/3(x+1)^(3/2)+c#