What is the integral of 2x(x^2+1)^{-5} ?

The integral of 2x(x^2+1)^{-5} is -1/(4(x^2+1)^4)+C

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integral of 2x(x^2+1)^{-5}

\int 2 x (x^{2} + 1)^{- 5} d x

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

Solution steps

\int 2 x (x^{2} + 1)^{- 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} + 1)^{- 5} d x

Simplify

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

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{a x}{y ^{2}})

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

t^{2} y^{^{''}} + 5 t y^{^{'}} + 4 y = 0

x \to 1 + lim (\frac{x - 1}{x - 1})

x \to 0 lim (\frac{2 sin ( \frac{x}{2} )}{x})

What is the integral of 2x(x^2+1)^{-5} ?
The integral of 2x(x^2+1)^{-5} is -1/(4(x^2+1)^4)+C