What is the integral of 2x(x^2+2)^{-5} ?

The integral of 2x(x^2+2)^{-5} is (x^2)/((x^2+2)^5)-5/(4(x^2+2)^4)+2/((x^2+2)^5)+C

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integral of 2x(x^2+2)^{-5}

\int 2 x (x^{2} + 2)^{- 5} d x

\frac{x ^{2}}{( x ^{2} + 2 ) ^{5}} - \frac{5}{4 ( x ^{2} + 2 ) ^{4}} + \frac{2}{( x ^{2} + 2 ) ^{5}} + C

Solution steps

\int 2 x (x^{2} + 2)^{- 5} d x

Take the constant out:

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

= 2 \cdot \int x (x^{2} + 2)^{- 5} d x

Simplify

x (x^{2} + 2)^{- 5} : \frac{x}{( x ^{2} + 2 ) ^{5}}

= 2 \cdot \int \frac{x}{( x ^{2} + 2 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{2 ( u + 2 ) ^{5}} d u

= 2 \cdot \int \frac{1}{2 ( u + 2 ) ^{5}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{( u + 2 ) ^{5}} d u

Apply Integration By Parts

: \frac{u}{( u + 2 ) ^{5}} - \int - \frac{5 u}{( u + 2 ) ^{6}} d u

= 2 \cdot \frac{1}{2} (\frac{u}{( u + 2 ) ^{5}} - \int - \frac{5 u}{( u + 2 ) ^{6}} d u)

\int - \frac{5 u}{( u + 2 ) ^{6}} d u = \frac{5}{4 ( u + 2 ) ^{4}} - \frac{2}{( u + 2 ) ^{5}}

= 2 \cdot \frac{1}{2} (\frac{u}{( u + 2 ) ^{5}} - (\frac{5}{4 ( u + 2 ) ^{4}} - \frac{2}{( u + 2 ) ^{5}}))

Substitute back

u = x^{2}

= 2 \cdot \frac{1}{2} (\frac{x ^{2}}{( x ^{2} + 2 ) ^{5}} - (\frac{5}{4 ( x ^{2} + 2 ) ^{4}} - \frac{2}{( x ^{2} + 2 ) ^{5}}))

Simplify

2 \cdot \frac{1}{2} (\frac{x ^{2}}{( x ^{2} + 2 ) ^{5}} - (\frac{5}{4 ( x ^{2} + 2 ) ^{4}} - \frac{2}{( x ^{2} + 2 ) ^{5}})) : \frac{x ^{2}}{( x ^{2} + 2 ) ^{5}} - \frac{5}{4 ( x ^{2} + 2 ) ^{4}} + \frac{2}{( x ^{2} + 2 ) ^{5}}

= \frac{x ^{2}}{( x ^{2} + 2 ) ^{5}} - \frac{5}{4 ( x ^{2} + 2 ) ^{4}} + \frac{2}{( x ^{2} + 2 ) ^{5}}

Add a constant to the solution

= \frac{x ^{2}}{( x ^{2} + 2 ) ^{5}} - \frac{5}{4 ( x ^{2} + 2 ) ^{4}} + \frac{2}{( x ^{2} + 2 ) ^{5}} + C

x^{5} y^{^{'}} + y^{5} = 0

x \to - 1 lim (e^{0})

\frac{d}{d x} (\frac{π}{( 3 x ) ^{2}})

\frac{d y}{d x} = 5 + y - 5 x + 8

\int sin^{3} (18 x) cos^{6} (18 x) d x

What is the integral of 2x(x^2+2)^{-5} ?
The integral of 2x(x^2+2)^{-5} is (x^2)/((x^2+2)^5)-5/(4(x^2+2)^4)+2/((x^2+2)^5)+C