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

The integral of (x^4+1)/(x(x^2+2)^2) is 3/8 ln|x^2+2|+1/4 ln(x)+1/(x^2+2)+1/(4(x^2+2))+C

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

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

\frac{3}{8} ln x^{2} + 2 + \frac{1}{4} ln (x) + \frac{1}{x ^{2} + 2} + \frac{1}{4 ( x ^{2} + 2 )} + C

Solution steps

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

Expand

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

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

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

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

Simplify

\frac{1}{2} ln x^{2} + 2 + \frac{1}{x ^{2} + 2} + \frac{1}{2} (\frac{1}{2} ln (x) - \frac{1}{4} ln x^{2} + 2 + \frac{1}{2 ( x ^{2} + 2 )}) : \frac{3}{8} ln x^{2} + 2 + \frac{1}{4} ln (x) + \frac{1}{x ^{2} + 2} + \frac{1}{4 ( x ^{2} + 2 )}

= \frac{3}{8} ln x^{2} + 2 + \frac{1}{4} ln (x) + \frac{1}{x ^{2} + 2} + \frac{1}{4 ( x ^{2} + 2 )}

Add a constant to the solution

= \frac{3}{8} ln x^{2} + 2 + \frac{1}{4} ln (x) + \frac{1}{x ^{2} + 2} + \frac{1}{4 ( x ^{2} + 2 )} + C

x \to π lim (ln ∣ cos (x) - 9 ∣)

- sec (\frac{π}{3}) + 2 sec^{3} (\frac{π}{3})

\int 20 - 0.01 x d x

5 e^{x} (cos (x) - sin (x))

\frac{d}{d x} (sin (x - 1))

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