What is the integral of 1/(x(1+x^2)^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x(1+x^2)^2)

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Take the partial fraction of

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

= \frac{1}{2} \cdot \int \frac{1}{u - 1} - \frac{1}{u} - \frac{1}{u ^{2}} 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

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

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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