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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2x(x^2+1)^{-4}

\int 2 x (x^{2} + 1)^{- 4} d x

- \frac{1}{3 ( x ^{2} + 1 ) ^{3}} + C

Solution steps

\int 2 x (x^{2} + 1)^{- 4} 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)^{- 4} d x

Simplify

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

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

Apply u-substitution

: \int \frac{1}{2 u ^{4}} d u

= 2 \cdot \int \frac{1}{2 u ^{4}} 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 ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 2 \cdot \frac{1}{2} (- \frac{1}{3 u ^{3}})

Substitute back

u = x^{2} + 1

= 2 \cdot \frac{1}{2} (- \frac{1}{3 ( x ^{2} + 1 ) ^{3}})

Simplify

2 \cdot \frac{1}{2} (- \frac{1}{3 ( x ^{2} + 1 ) ^{3}}) : - \frac{1}{3 ( x ^{2} + 1 ) ^{3}}

= - \frac{1}{3 ( x ^{2} + 1 ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( x ^{2} + 1 ) ^{3}} + C

\frac{7}{s ^{3} + 2 s ^{2} + 9 s + 18}

\int \frac{e ^{x}}{e ^{- 6 x}} d x

\frac{d}{d x} (\frac{5 x}{9 x ^{2} + 1})

y^{^{'}} = y - k

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

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