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 $2/3(x+1)^(3/2)+c$

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

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