What is the integral of (x^4+1)/(x(x^2+1)^2) ?

The integral of (x^4+1)/(x(x^2+1)^2) is ln|x|+1/(x^2+1)+C

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integral of (x^4+1)/(x(x^2+1)^2)

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

ln ∣ x ∣ + \frac{1}{x ^{2} + 1} + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4} + 1}{x ( x ^{2} + 1 ) ^{2}} : \frac{1}{x} - \frac{2 x}{( x ^{2} + 1 ) ^{2}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - (- \frac{1}{x ^{2} + 1})

Simplify

= ln ∣ x ∣ + \frac{1}{x ^{2} + 1}

Add a constant to the solution

= ln ∣ x ∣ + \frac{1}{x ^{2} + 1} + C

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

e^{4 x}

\int_{0}^{5} \frac{1}{x ^{\frac{15}{16}}} d x

y = 6 x^{2} ln (x), y = 24 ln (x)

\frac{\partial}{\partial z} (x z - e^{x} sin (y))

What is the integral of (x^4+1)/(x(x^2+1)^2) ?
The integral of (x^4+1)/(x(x^2+1)^2) is ln|x|+1/(x^2+1)+C