What is the integral of (x^3+4x+3)/(x^4+5x^2+4) ?

The integral of (x^3+4x+3)/(x^4+5x^2+4) is 1/2 ln|x^2+1|+arctan(x)-1/2 arctan(x/2)+C

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integral of (x^3+4x+3)/(x^4+5x^2+4)

\int \frac{x ^{3} + 4 x + 3}{x ^{4} + 5 x ^{2} + 4} d x

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

Solution steps

\int \frac{x ^{3} + 4 x + 3}{x ^{4} + 5 x ^{2} + 4} d x

Take the partial fraction of

\frac{x ^{3} + 4 x + 3}{x ^{4} + 5 x ^{2} + 4} : \frac{x + 1}{x ^{2} + 1} - \frac{1}{x ^{2} + 4}

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

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

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

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

Add a constant to the solution

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

\int (sec (5 x) + csc (5 x))^{2} d x

n = 8 \sum \infty cos (\frac{1}{n ^{3}})

d er i v a t i v e f (x) = sin (3 ln (x))

d er i v a t i v e y = 11 x + 11 x + 11 x

\int x^{2} \cdot cos (x) d x

What is the integral of (x^3+4x+3)/(x^4+5x^2+4) ?
The integral of (x^3+4x+3)/(x^4+5x^2+4) is 1/2 ln|x^2+1|+arctan(x)-1/2 arctan(x/2)+C