What is the integral of (x^4+1)/(x^5+4x^3) ?

The integral of (x^4+1)/(x^5+4x^3) is -1/16 ln|x|-1/(8x^2)+17/32 ln|x^2+4|+C

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integral of (x^4+1)/(x^5+4x^3)

\int \frac{x ^{4} + 1}{x ^{5} + 4 x ^{3}} d x

- \frac{1}{16} ln ∣ x ∣ - \frac{1}{8 x ^{2}} + \frac{17}{32} ln x^{2} + 4 + C

Solution steps

\int \frac{x ^{4} + 1}{x ^{5} + 4 x ^{3}} d x

Take the partial fraction of

\frac{x ^{4} + 1}{x ^{5} + 4 x ^{3}} : - \frac{1}{16 x} + \frac{1}{4 x ^{3}} + \frac{17 x}{16 ( x ^{2} + 4 )}

= \int - \frac{1}{16 x} + \frac{1}{4 x ^{3}} + \frac{17 x}{16 ( x ^{2} + 4 )} 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}{16 x} d x + \int \frac{1}{4 x ^{3}} d x + \int \frac{17 x}{16 ( x ^{2} + 4 )} d x

\int \frac{1}{16 x} d x = \frac{1}{16} ln ∣ x ∣

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

\int \frac{17 x}{16 ( x ^{2} + 4 )} d x = \frac{17}{32} ln x^{2} + 4

= - \frac{1}{16} ln ∣ x ∣ - \frac{1}{8 x ^{2}} + \frac{17}{32} ln x^{2} + 4

Add a constant to the solution

= - \frac{1}{16} ln ∣ x ∣ - \frac{1}{8 x ^{2}} + \frac{17}{32} ln x^{2} + 4 + C

n = 1 \sum \infty (- 1)^{n - 1}

\frac{\partial}{\partial x} (2 x y - x^{3} - y^{2})

\int t (e^{- s t}) d t

\frac{3}{2} t + 2 e^{- t} - \frac{1}{4} e^{- 2 t} - \frac{7}{4}

\int \frac{3 x + 1}{( x - 2 ) ( x + 3 ) ^{2}} d x

What is the integral of (x^4+1)/(x^5+4x^3) ?
The integral of (x^4+1)/(x^5+4x^3) is -1/16 ln|x|-1/(8x^2)+17/32 ln|x^2+4|+C