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

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

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

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

ln ∣ x ∣ - ln x^{2} + 1 + C

Solution steps

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

Expand

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

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

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

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

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

Add similar elements:

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

= ln ∣ x ∣ - ln x^{2} + 1

Add a constant to the solution

= ln ∣ x ∣ - ln x^{2} + 1 + C

\int e^{x} + x^{2} d x

x \to 0 lim (0)

\frac{d}{d x} (3 x^{7})

\int_{4}^{\infty} x e^{- x} d x

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