What is the integral of (x^3+3x+4)/(x^4+16x^2) ?

The integral of (x^3+3x+4)/(x^4+16x^2) is 13/32 ln|x^2+16|+3/16 ln(x)-1/16 arctan(x/4)-1/(4x)+C

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integral of (x^3+3x+4)/(x^4+16x^2)

\int \frac{x ^{3} + 3 x + 4}{x ^{4} + 16 x ^{2}} d x

\frac{13}{32} ln x^{2} + 16 + \frac{3}{16} ln (x) - \frac{1}{16} arctan (\frac{x}{4}) - \frac{1}{4 x} + C

Solution steps

\int \frac{x ^{3} + 3 x + 4}{x ^{4} + 16 x ^{2}} d x

Expand

\frac{x ^{3} + 3 x + 4}{x ^{4} + 16 x ^{2}} : \frac{x ^{3}}{x ^{4} + 16 x ^{2}} + \frac{3 x}{x ^{4} + 16 x ^{2}} + \frac{4}{x ^{4} + 16 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{x ^{3}}{x ^{4} + 16 x ^{2}} d x + \int \frac{3 x}{x ^{4} + 16 x ^{2}} d x + \int \frac{4}{x ^{4} + 16 x ^{2}} d x

\int \frac{x ^{3}}{x ^{4} + 16 x ^{2}} d x = \frac{1}{2} ln x^{2} + 16

\int \frac{3 x}{x ^{4} + 16 x ^{2}} d x = 3 (\frac{1}{16} ln (x) - \frac{1}{32} ln x^{2} + 16)

\int \frac{4}{x ^{4} + 16 x ^{2}} d x = 4 (- \frac{1}{16 x} - \frac{1}{64} arctan (\frac{x}{4}))

= \frac{1}{2} ln x^{2} + 16 + 3 (\frac{1}{16} ln (x) - \frac{1}{32} ln x^{2} + 16) + 4 (- \frac{1}{16 x} - \frac{1}{64} arctan (\frac{x}{4}))

Simplify

\frac{1}{2} ln x^{2} + 16 + 3 (\frac{1}{16} ln (x) - \frac{1}{32} ln x^{2} + 16) + 4 (- \frac{1}{16 x} - \frac{1}{64} arctan (\frac{x}{4})) : \frac{13}{32} ln x^{2} + 16 + \frac{3}{16} ln (x) - \frac{1}{16} arctan (\frac{x}{4}) - \frac{1}{4 x}

= \frac{13}{32} ln x^{2} + 16 + \frac{3}{16} ln (x) - \frac{1}{16} arctan (\frac{x}{4}) - \frac{1}{4 x}

Add a constant to the solution

= \frac{13}{32} ln x^{2} + 16 + \frac{3}{16} ln (x) - \frac{1}{16} arctan (\frac{x}{4}) - \frac{1}{4 x} + C

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

x \to - 8 lim (- \frac{3}{7} x + \frac{10}{11})

\frac{d}{d x} (\frac{7}{x - 1})

f (x) = x^{2}, (2, 4)

What is the integral of (x^3+3x+4)/(x^4+16x^2) ?
The integral of (x^3+3x+4)/(x^4+16x^2) is 13/32 ln|x^2+16|+3/16 ln(x)-1/16 arctan(x/4)-1/(4x)+C