How do you integrate $\int \frac{x}{\sqrt{x + 1}}$ $int x/sqrt(x+1)$ using substitution?

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How do you integrate $\int \frac{x}{\sqrt{x + 1}}$ $int x/sqrt(x+1)$ using substitution?

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Answer 1 · 2016-10-24T12:38:53

$y = \frac{2 {(x + 1)}^{\frac{3}{2}}}{3} - 2 \sqrt{x + 1} + c$ $y=(2(x+1)^(3/2))/3-2sqrt(x+1)+c$

Explanation:

so let us take,

$u = x + 1$ $u=x+1$

so we differentiate this to find that

$\frac{d u}{d x} = 1$ $(du)/(dx)=1$

and $x = u - 1$ $x=u-1$

so let's rewrite our original integral

$\int (\frac{x}{\sqrt{x + 1}}) d x$ $int(x/sqrt(x+1))dx$

$= \int (\frac{u - 1}{\sqrt{u}}) (1) d x$ $=int((u-1)/sqrt(u))(1)dx$

$= \int (\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}) \frac{d u}{d x} d x$ $=int(u/sqrt(u)-1/sqrt(u))(du)/(dx)dx$

$= \int (\sqrt{u} - \frac{1}{\sqrt{u}}) (d u)$ $=int(sqrt(u)-1/sqrt(u))(du)$

$= \frac{2 u^{\frac{3}{2}}}{3} - 2 \sqrt{u} + c$ $=(2u^(3/2))/3-2sqrt(u)+c$

we can now sub in the $x + 1 for the u's$ $x+1 " for the u's"$ if we need,

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

Answer 2 · 2016-10-24T12:50:46

The integral is $= \frac{2}{3} \sqrt{x + 1} (x - 2) + C$ $=2/3sqrt(x+1)(x-2)+C$

Explanation:

Let $u = x + 1$ $u=x+1$ and $x = u - 1$ $x=u-1$

the $d u = d x$ $du=dx$
So $\int \frac{x d x}{\sqrt{x + 1}} = \int \frac{(u - 1) d u}{u^{\frac{1}{2}}}$ $int(xdx)/sqrt(x+1)=int((u-1)du)/u^(1/2)$

$= \int (u^{\frac{1}{2}} - u^{- \frac{1}{2}}) d u$ $=int(u^(1/2)-u^(-1/2))du$

$= \frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{1}{2}}}{\frac{1}{2}}$ $=u^(3/2)/(3/2)-u^(1/2)/(1/2)$

$= \frac{2}{3} u \cdot \sqrt{u} - 2 \sqrt{u}$ $=2/3u*sqrtu-2sqrtu$
$= 2 \cdot \sqrt{u} (\frac{1}{3} u - 1)$ $=2*sqrtu(1/3u-1)$
$= \frac{2}{3} \sqrt{u} (u - 3)$ $=2/3sqrtu(u-3)$

Replacing u by (x+1)

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

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

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