What is the integral of 2x(x^2+4)^{-3} ?

The integral of 2x(x^2+4)^{-3} is -1/(2(x^2+4)^2)+C

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integral of 2x(x^2+4)^{-3}

\int 2 x (x^{2} + 4)^{- 3} d x

- \frac{1}{2 ( x ^{2} + 4 ) ^{2}} + C

Solution steps

\int 2 x (x^{2} + 4)^{- 3} 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} + 4)^{- 3} d x

Simplify

x (x^{2} + 4)^{- 3} : \frac{x}{( x ^{2} + 4 ) ^{3}}

= 2 \cdot \int \frac{x}{( x ^{2} + 4 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= 2 \cdot \int \frac{1}{2 u ^{3}} 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 ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = x^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 2x(x^2+4)^{-3} ?
The integral of 2x(x^2+4)^{-3} is -1/(2(x^2+4)^2)+C