Question

$\displaystyle\sf x = 3+2\sqrt{2}$
$\displaystyle\sf \sqrt{x}-\dfrac{1}{\sqrt{x}}$??
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Answer 1

$\huge\frak\red{Ꭺnsᗯer}$

$\begin{gathered} \sf \: given \: \: \boxed{ \sf u = 1 - {r}^{3} } \\ \ \\ \displaystyle \sf\int \limits \dfrac{ {9r}^{2} }{ \sqrt{1 - {r}^{3} } } \, dr \\ \\ \\ \sf \ \frac{d(1 - {r}^{3} )}{dr} = \frac{du}{dr} \\ \\ \\ \sf \: - 3 {r}^{3 - 1} = \frac{du}{dr} \\ \\ \\ \sf \: - 3 {r}^{2} = \frac{du}{dr} \\ \\ \\ \sf \: 3 \int \frac{ {3r}^{2} }{ \sqrt{1 - {r}^{3} } } dr \\ \\ \\ \sf \: - 3 \int{ \frac{ du}{ \sqrt{u} } } \\ \\ \\ \sf \: - 3 \int \: \frac{du}{ {u}^{ \frac{1}{2} } } \\ \\ \ \\ \sf \: - 3 \int( {u}^{ \frac{ - 1}{2}) } du \\ \\ \\ \sf \: - 3( \frac{ {u}^{? \frac{ - 1 + 2}{2} } }{ - \frac{1}{2} + 2} ) + c \\ \\ \\ \sf \: - 3( \frac{ {u}^{ \frac{1}{2} } }{ \frac{1}{2} } ) + c \\ \\ \\ \sf \: - 3(2 {u}^{ \frac{1}{2} } )+ c \\ \\ \\ \sf \: - 6 \sqrt{u} + c \\ \\ \sf \: put \: value \: of \: u : - \\ \\ \sf \: - 6 \sqrt{1 - {r}^{3} } + c\end{gathered}$

Answer 2

Answer:

Hope it is helpful to you

Explanation:

In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. In an algebraic equation, the left-hand side is equal to the right-hand side.

There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.