What is the integral of 2x(x^2+1)^{-6} ?

The integral of 2x(x^2+1)^{-6} is -1/(5(x^2+1)^5)+C

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integral of 2x(x^2+1)^{-6}

\int 2 x (x^{2} + 1)^{- 6} d x

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

Solution steps

\int 2 x (x^{2} + 1)^{- 6} 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)^{- 6} d x

Simplify

x (x^{2} + 1)^{- 6} : \frac{x}{( x ^{2} + 1 ) ^{6}}

= 2 \cdot \int \frac{x}{( x ^{2} + 1 ) ^{6}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{6}} d u

= 2 \cdot \int \frac{1}{2 u ^{6}} 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 ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

f (t) = \frac{1}{( t ^{2} + 3 t + 1 ) ^{\frac{5}{2}}}

\int 1 - e^{x} d x

x \to - \infty lim (x^{3} - x)

x \to \infty lim (x^{- 6})

\int - x \cdot e^{3 x} d x

What is the integral of 2x(x^2+1)^{-6} ?
The integral of 2x(x^2+1)^{-6} is -1/(5(x^2+1)^5)+C