Question

Integrate using integration by parts. ? x?(x+2) dx = ? x(x+2)^{1/2} dx ? udv = uv - ? vdu u = x ? du = dx dv = (x+2)^{1/2} dx v = 2/3(x+2)^{3/2} ? u^r du = u^{r+1}/(r+1) + C, r ? -1 ? e^u du = e^u + C ? 1/u du = ln|u| + C ? (x+2)^{1/2} dx u = x + 2 du = dx = (x+2)^{3/2} / (3/2) + C

          Integrate using integration by parts.
? x?(x+2) dx = ? x(x+2)^{1/2} dx
? udv = uv - ? vdu
u = x ? du = dx
dv = (x+2)^{1/2} dx
v = 2/3(x+2)^{3/2}
? u^r du = u^{r+1}/(r+1) + C, r ? -1
? e^u du = e^u + C
? 1/u du = ln|u| + C
? (x+2)^{1/2} dx
u = x + 2
du = dx
= (x+2)^{3/2} / (3/2) + C

Integrate using integration by parts.
? x?(x+2) dx = ? x(x+2)^1/2 dx
? udv = uv - ? vdu
u = x ? du = dx
dv = (x+2)^1/2 dx
v = 2/3(x+2)^3/2
? u^r du = u^r+1/(r+1) + C, r ? -1
? e^u du = e^u + C
? 1/u du = ln|u| + C
? (x+2)^1/2 dx
u = x + 2
du = dx
= (x+2)^3/2 / (3/2) + C

Added by Noelia B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/23/2023

Step 1

We need to choose $u$ and $dv$. Let's choose: $$u = x, \quad dv = \sqrt{x} + 2 \, dx$$ Now, we need to find $du$ and $v$. We have: $$du = dx, \quad v = \int (\sqrt{x} + 2) \, dx$$ To find $v$, we need to integrate $\sqrt{x} + 2$ with respect to $x$: $$v = \int Show more…

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