What is the integral of (7x^3+4x+4x^2+1)/(x^4+x^2) ?

The integral of (7x^3+4x+4x^2+1)/(x^4+x^2) is 3/2 ln|x^2+1|+4ln(x)+3arctan(x)-1/x+C

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integral of (7x^3+4x+4x^2+1)/(x^4+x^2)

\int \frac{7 x ^{3} + 4 x + 4 x ^{2} + 1}{x ^{4} + x ^{2}} d x

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

Solution steps

\int \frac{7 x ^{3} + 4 x + 4 x ^{2} + 1}{x ^{4} + x ^{2}} d x

Expand

\frac{7 x ^{3} + 4 x + 4 x ^{2} + 1}{x ^{4} + x ^{2}} : \frac{7 x ^{3}}{x ^{4} + x ^{2}} + \frac{4 x}{x ^{4} + x ^{2}} + \frac{4 x ^{2}}{x ^{4} + x ^{2}} + \frac{1}{x ^{4} + x ^{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{7 x ^{3}}{x ^{4} + x ^{2}} d x + \int \frac{4 x}{x ^{4} + x ^{2}} d x + \int \frac{4 x ^{2}}{x ^{4} + x ^{2}} d x + \int \frac{1}{x ^{4} + x ^{2}} d x

\int \frac{7 x ^{3}}{x ^{4} + x ^{2}} d x = \frac{7}{2} ln x^{2} + 1

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

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

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

= \frac{7}{2} ln x^{2} + 1 + 4 (- \frac{1}{2} ln x^{2} + 1 + ln (x)) + 4 arctan (x) - \frac{1}{x} - arctan (x)

Simplify

\frac{7}{2} ln x^{2} + 1 + 4 (- \frac{1}{2} ln x^{2} + 1 + ln (x)) + 4 arctan (x) - \frac{1}{x} - arctan (x) : \frac{3}{2} ln x^{2} + 1 + 4 ln (x) + 3 arctan (x) - \frac{1}{x}

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

Add a constant to the solution

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

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

\int \frac{4 x}{x - 6} d x

\frac{\partial}{\partial z} (x y + z)

y^{^{'}} = 1 + 4 y^{2}, y (1) = 0

\int_{1}^{2} x^{3} x^{4} + 5 d x

What is the integral of (7x^3+4x+4x^2+1)/(x^4+x^2) ?
The integral of (7x^3+4x+4x^2+1)/(x^4+x^2) is 3/2 ln|x^2+1|+4ln(x)+3arctan(x)-1/x+C