Question #6f533

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Question #6f533

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Answer 1 · 2016-12-15T01:18:52

$\int x \sqrt{2 x + 1} d x = {(2 x + 1)}^{\frac{3}{2}} (\frac{1}{5} x - \frac{1}{15}) + C$ $intxsqrt(2x+1)dx = (2x+1)^(3/2)(1/5x-1/15)+C$

Explanation:

Using the integration by parts formula

$\int u d v = u v - \int v d u$ $intudv = uv-intvdu$

Let
$u = x \Rightarrow d u = d x$ $u = x => du = dx$
$d v = {(2 x + 1)}^{\frac{1}{2}} d x \Rightarrow v = \frac{1}{3} {(2 x + 1)}^{\frac{3}{2}}$ $dv = (2x+1)^(1/2)dx => v = 1/3(2x+1)^(3/2)$

$\int x \sqrt{2 x + 1} d x = \int u d v$ $intxsqrt(2x+1)dx = intudv$

$= u v - \int v d u$ $=uv - intvdu$

$= \frac{1}{3} x {(2 x + 1)}^{\frac{3}{2}} - \frac{1}{3} \int {(2 x + 1)}^{\frac{3}{2}} d x$ $=1/3x(2x+1)^(3/2)-1/3int(2x+1)^(3/2)dx$

$= \frac{1}{3} x {(2 x + 1)}^{\frac{3}{2}} - \frac{1}{3} [\frac{1}{5} {(2 x + 1)}^{\frac{5}{2}}] + C$ $=1/3x(2x+1)^(3/2)-1/3[1/5(2x+1)^(5/2)]+C$

$= \frac{1}{3} x {(2 x + 1)}^{\frac{3}{2}} - \frac{1}{15} {(2 x + 1)}^{\frac{5}{2}} + C$ $=1/3x(2x+1)^(3/2)-1/15(2x+1)^(5/2)+C$

$= \frac{1}{3} {(2 x + 1)}^{\frac{3}{2}} [x - \frac{1}{5} (2 x + 1)] + C$ $=1/3(2x+1)^(3/2)[x-1/5(2x+1)]+C$

$= {(2 x + 1)}^{\frac{3}{2}} (\frac{1}{5} x - \frac{1}{15}) + C$ $=(2x+1)^(3/2)(1/5x-1/15)+C$

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Question #6f533

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