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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^{-1)+x^{-3}}

\int \frac{1}{x ^{- 1} + x ^{- 3}} d x

- ln ∣ x ∣ + \frac{x ^{2}}{2} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 + C

Solution steps

\int \frac{1}{x ^{- 1} + x ^{- 3}} d x

Apply u-substitution

: \int - \frac{1}{u ^{3} ( u ^{2} + 1 )} d u

= \int - \frac{1}{u ^{3} ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u ^{3} ( u ^{2} + 1 )} d u

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int \frac{1}{u} d u + \int \frac{1}{u ^{3}} d u + \int \frac{u}{u ^{2} + 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

= - (- ln ∣ u ∣ - \frac{1}{2 u ^{2}} + \frac{1}{2} ln u^{2} + 1)

Substitute back

u = x^{- 1}

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

Simplify

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

= - ln ∣ x ∣ + \frac{x ^{2}}{2} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1

Add a constant to the solution

= - ln ∣ x ∣ + \frac{x ^{2}}{2} - \frac{1}{2} ln \frac{1}{x ^{2}} + 1 + C

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

y = - 2 x^{3}, (2, - 16)

x \to 3 lim (\frac{∣ x + 3 ∣}{x - 3})

\int y + 1 d y

\int_{- 5}^{0} 4 + 25 - x^{2} d x

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