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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} ((x^{2} + 8)^{5})

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

\frac{x ^{2} + 6}{5 x - 2}

\int e^{(y^{2})} d y

\int_{0}^{1} (x - x) d x

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