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

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

Popular >

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1

Add a constant to the solution

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

y^{^{''}} + 6 y^{^{'}} + 11 y = 0, y (0) = - 1, y^{^{'}} (0) = - 4

\frac{d}{d x} (\frac{27}{x ^{2} + 9})

\int_{0}^{7} \frac{2 t}{( t - 8 ) ^{2}} d t

\frac{\partial}{\partial x} (\frac{2 πk t}{ln ( \frac{r}{x} )})

\frac{d}{d t} (t \cdot sin (t))

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