Question #362bd

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Question #362bd

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Answer 1 · 2016-11-18T12:51:15

$2 {log}_{e} (\sqrt{x} + 1)$ $2log_e(sqrtx+1)$

Explanation:

Note that $\frac{1}{x + \sqrt{x}} = \frac{1}{\sqrt{x} (\sqrt{x} + 1)}$ $1/(x+sqrt(x))=1/(sqrtx(sqrtx+1))$ and also that
$\frac{d}{d x} (\sqrt{x} + 1) = \frac{1}{2} (\frac{1}{\sqrt{x}})$ $d/dx(sqrtx+1)=1/2(1/sqrtx)$ so

$\frac{1}{\sqrt{x} (\sqrt{x} + 1)} = 2 \frac{\frac{d}{d x} (\sqrt{x} + 1)}{\sqrt{x} + 1}$ $1/(sqrtx(sqrtx+1))=2(d/dx(sqrtx+1))/(sqrtx+1)$ so

$\int \frac{d x}{x + \sqrt{x}} = \int \frac{1}{x + \sqrt{x}} d x = \int 2 \frac{\frac{d}{d x} (\sqrt{x} + 1)}{\sqrt{x} + 1} d x = 2 {log}_{e} (\sqrt{x} + 1)$ $intdx/(x+sqrt(x))=int1/(x+sqrt(x))dx=int2(d/dx(sqrtx+1))/(sqrtx+1)dx=2log_e(sqrtx+1)$

Answer 2 · 2016-11-18T12:53:45

The answer is $= 2 ln (1 + \sqrt{x}) + C$ $=2ln(1+sqrtx)+C$

Explanation:

The two writing represent the same integral.

Let $I = \int \frac{d x}{x + \sqrt{x}}$ $I=intdx/(x+sqrtx)$

Multiply numerator and denominator by $x - \sqrt{x}$ $x-sqrtx$

$I = \int \frac{(x - \sqrt{x}) d x}{(x + \sqrt{x}) (x - \sqrt{x})}$ $I=int((x-sqrtx)dx)/((x+sqrtx)(x-sqrtx))$

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

Solving by substitution

Let $u = \sqrt{x}$ $u=sqrtx$ ;
$u^{2} = x$ $u^2=x$

$u^{4} = x^{2}$ $u^4 =x^2$

$d u = \frac{d x}{2 \sqrt{x}}$ $du=dx/(2sqrtx)$

$d x = 2 u d u$ $dx=2udu$

$I = \int (u^{2} - u) \frac{2 u d u}{u^{4} - u^{2}}$ $I=int(u^2-u)(2udu)/(u^4-u^2)$

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

$=2int(cancel(u-1)du)/((u+1)cancel(u-1))$

$=2ln(u+1)$

$=2ln(1+sqrtx)+C$

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Question #362bd

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