What is the integral of 2x(x^2+1)^{-7} ?

The integral of 2x(x^2+1)^{-7} is -1/(6(x^2+1)^6)+C

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integral of 2x(x^2+1)^{-7}

\int 2 x (x^{2} + 1)^{- 7} d x

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

Solution steps

\int 2 x (x^{2} + 1)^{- 7} 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)^{- 7} d x

Simplify

x (x^{2} + 1)^{- 7} : \frac{x}{( x ^{2} + 1 ) ^{7}}

= 2 \cdot \int \frac{x}{( x ^{2} + 1 ) ^{7}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{7}} d u

= 2 \cdot \int \frac{1}{2 u ^{7}} 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 ^{7}} d u

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= 2 \cdot \frac{1}{2} (- \frac{1}{6 u ^{6}})

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{2}, [0, 3]

\int_{- 1}^{3} (x^{2} + 6 x + 3) d x

\frac{d ^{3}}{d x ^{3}} (x^{4} - 10 x^{3})

\frac{\partial}{\partial x} (\frac{1}{y + z})

f (x) = \frac{5}{x ^{5}}

What is the integral of 2x(x^2+1)^{-7} ?
The integral of 2x(x^2+1)^{-7} is -1/(6(x^2+1)^6)+C