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

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

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

ln ∣ x ∣ - ln 1 + x^{2} + C

Solution steps

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

Expand

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

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

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

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

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

Add similar elements:

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

= ln ∣ x ∣ - ln 1 + x^{2}

Add a constant to the solution

= ln ∣ x ∣ - ln 1 + x^{2} + C

\int \frac{8 x ^{3} + 1}{4 x ^{2} - 2 x + 1} d x

\int (\frac{3}{5} x^{8} - \frac{4}{3} x^{3})^{2} d x

\int ln^{2} (5 x^{3}) d x

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

\int e^{- y^{3}} d y

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