What is the integral of (3x^3+x)/(x^4+2x^2+1) ?

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

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integral of (3x^3+x)/(x^4+2x^2+1)

\int \frac{3 x ^{3} + x}{x ^{4} + 2 x ^{2} + 1} d x

\frac{3}{2} ln x^{2} + 1 + \frac{1}{x ^{2} + 1} + C

Solution steps

\int \frac{3 x ^{3} + x}{x ^{4} + 2 x ^{2} + 1} d x

Expand

\frac{3 x ^{3} + x}{x ^{4} + 2 x ^{2} + 1} : \frac{3 x ^{3}}{x ^{4} + 2 x ^{2} + 1} + \frac{x}{x ^{4} + 2 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{3 x ^{3}}{x ^{4} + 2 x ^{2} + 1} d x + \int \frac{x}{x ^{4} + 2 x ^{2} + 1} d x

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

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

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

Simplify

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

= \frac{3}{2} ln x^{2} + 1 + \frac{1}{x ^{2} + 1}

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 1 + \frac{1}{x ^{2} + 1} + C

\frac{1}{( s ^{2} + 3 ) ( s ^{2} + 1 )}

\int \frac{sec ^{2} ( x )}{csc ( x )} d x

\frac{d}{d x} (lo g_{10} (4 x))

\int \frac{1}{t - \frac{1}{2}} d t

e^{- y} sin (x) + y^{^{'}} = 0

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