Question #800f5

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Question #800f5

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Answer 1 · 2017-06-04T09:25:31

$\int {(x^{\frac{1}{2}} - x^{- \frac{1}{2}})}^{2} d x = \int (x - 2 + x^{- 1}) d x$ $\int(x^(1/2)-x^(-1/2))^2"d"x = \int(x-2+x^-1)"d"x$
$\int {(x^{\frac{1}{2}} - x^{- \frac{1}{2}})}^{2} d x = \frac{x^{2}}{2} - 2 x + ln | x | + C$ $\int(x^(1/2)-x^(-1/2))^2"d"x = x^2/2-2x+ln\abs(x)+C$

Answer 2 · 2017-06-04T10:45:19

$\frac{x^{2}}{2} - 2 x + ln x + C .$ $x^2/2-2x+lnx+C.$

Explanation:

Let, $I = \int {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} d x .$ $I=int(sqrtx-1/sqrtx)^2dx.$

Knowing that, $(1) : \int y^{n} d y = \frac{y^{n + 1}}{n + 1} + c, n \neq - 1, and,$ $(1) : inty^ndy=y^(n+1)/(n+1)+c, n!=-1, and,$

$(2) : int1/xdx=ln|x|+c',$ we have,

$I=int(sqrtx-1/sqrtx)^2dx,$

$=int(x-2+1/x)dx,............[because," squaring]"$

$=intx^1dx-2intx^0dx+int1/xdx,$

$=x^(1+1)/2-2*x^(0+1)/(0+1)+ln|x|,$

$rArr I=x^2/2-2x+ln|x|+C.$

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Question #800f5

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