Question #56aaf

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Question #56aaf

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Answer 1 · 2017-02-14T17:03:30

$\int (x - 1) \sqrt{2 + x} d x = (x - 1) \frac{2}{3} {(2 + x)}^{\frac{3}{2}} + \frac{4}{15} {(2 + x)}^{\frac{5}{2}} + C$ $int(x - 1)sqrt(2 + x)dx = (x - 1)2/3(2 + x)^(3/2) + 4/15(2 + x)^(5/2) + C$

Explanation:

Let $u = x - 1$ $u = x - 1$ and $d v = \sqrt{2 + x} d x$ $dv = sqrt(2 + x)dx$ . By the product rule, $d u = d x$ $du = dx$ . To integrate $d v$ $dv$ however, we will need to make a substitution.

Let $n = 2 + x$ $n = 2 + x$ . Then $d n = d x$ $dn = dx$ .

$d v = \sqrt{n} d n$ $dv = sqrt(n)dn$

$\int d v = \int \sqrt{n} d n$ $intdv = intsqrt(n)dn$

$v = \frac{2}{3} n^{\frac{3}{2}} \to$ $v = 2/3n^(3/2) ->$ We can add $C$ $C$ later on

$v = \frac{2}{3} {(2 + x)}^{\frac{3}{2}}$ $v = 2/3(2 + x)^(3/2)$

Now use the integration by parts formula.

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

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

You're going to need substitution to evaluate this integral. Let $m = 2 + x$ $m = 2 + x$ . Then $d m = d x$ $dm = dx$ .

$\int \frac{2}{3} {(2 + x)}^{\frac{3}{2}} d x = \int \frac{2}{3} m^{\frac{3}{2}} d m$ $int2/3(2 + x)^(3/2)dx = int2/3m^(3/2)dm$

Now use $\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $intx^ndx = x^(n + 1)/(n+1) + C$ , as we did above.

$\int \frac{2}{3} {(2 + x)}^{\frac{3}{2}} d x = \frac{4}{15} m^{\frac{5}{2}}$ $int2/3(2 + x)^(3/2)dx = 4/15m^(5/2)$

Reverse the substitution:

$\int \frac{2}{3} {(2 + x)}^{\frac{3}{2}} d x = - \frac{4}{15} {(2 + x)}^{\frac{5}{2}}$ $int2/3(2 + x)^(3/2)dx = -4/15(2 + x)^(5/2)$

We can now put everything together.

$\int (x - 1) \sqrt{2 + x} d x = (x - 1) \frac{2}{3} {(2 + x)}^{\frac{3}{2}} + \frac{4}{15} {(2 + x)}^{\frac{5}{2}} + C$ $int(x - 1)sqrt(2 + x)dx = (x - 1)2/3(2 + x)^(3/2) + 4/15(2 + x)^(5/2) + C$

Hopefully this helps!

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