What is the integral of (-2)/(sqrt(k-2)) ?

The integral of (-2)/(sqrt(k-2)) is -4sqrt(k-2)+C

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integral of (-2)/(sqrt(k-2))

integral

\frac{- 2}{k - 2}

- 4 k - 2 + C

Solution steps

\int \frac{- 2}{k - 2} d k

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 2 \cdot \int \frac{1}{k - 2} d k

Apply u-substitution

= - 2 \cdot \int 2 d u

Integral of a constant:

\int a d x = a x

= - 2 \cdot 2 u

Substitute back

u = k - 2

= - 2 \cdot 2 k - 2

Simplify

= - 4 k - 2

Add a constant to the solution

= - 4 k - 2 + C

in t e g r a l \frac{1}{x ( x + 1 ) ^{\frac{1}{2}}}

in t e g r a l \frac{x ^{2}}{( x - 2 ) ( x ^{2} + 1 )}

in t e g r a l y = \frac{x ^{2} - cos ( x ) + 1}{x}

in t e g r a l \frac{0.001}{( 0.001 x + 1 ) ^{2}}

in t e g r a l x - 3

What is the integral of (-2)/(sqrt(k-2)) ?
The integral of (-2)/(sqrt(k-2)) is -4sqrt(k-2)+C