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

The integral of (3x(3x+2)+4)/(x^4+2x^2) is 3ln|x|-2/x-3/2 ln|x^2+2|+7/(sqrt(2))arctan(x/(sqrt(2)))+C

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

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

3 ln ∣ x ∣ - \frac{2}{x} - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2}) + C

Solution steps

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

Take the partial fraction of

\frac{3 x ( 3 x + 2 ) + 4}{x ^{4} + 2 x ^{2}} : \frac{3}{x} + \frac{2}{x ^{2}} + \frac{- 3 x + 7}{x ^{2} + 2}

= \int \frac{3}{x} + \frac{2}{x ^{2}} + \frac{- 3 x + 7}{x ^{2} + 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{3}{x} d x + \int \frac{2}{x ^{2}} d x + \int \frac{- 3 x + 7}{x ^{2} + 2} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{- 3 x + 7}{x ^{2} + 2} d x = - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2})

= 3 ln ∣ x ∣ - \frac{2}{x} - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2})

Add a constant to the solution

= 3 ln ∣ x ∣ - \frac{2}{x} - \frac{3}{2} ln x^{2} + 2 + \frac{7}{2} arctan (\frac{x}{2}) + C

\int \frac{1}{( arctan ( \frac{y}{7} ) ( 49 + y ^{2} ) )} d y

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

\int \frac{5 x}{8} d x

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

What is the integral of (3x(3x+2)+4)/(x^4+2x^2) ?
The integral of (3x(3x+2)+4)/(x^4+2x^2) is 3ln|x|-2/x-3/2 ln|x^2+2|+7/(sqrt(2))arctan(x/(sqrt(2)))+C