What is the integral of (x^3+x^2+8x+7)/(6-x^2-x^4) ?

The integral of (x^3+x^2+8x+7)/(6-x^2-x^4) is

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integral of (x^3+x^2+8x+7)/(6-x^2-x^4)

\int \frac{x ^{3} + x ^{2} + 8 x + 7}{6 - x ^{2} - x ^{4}} d x

\frac{9 - 10 2}{10 2} ln x + 2 - \frac{10 2 + 9}{10 2} ln x - 2 + \frac{1}{2} ln x^{2} + 3 + \frac{4}{5 3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{x ^{3} + x ^{2} + 8 x + 7}{6 - x ^{2} - x ^{4}} d x

Take the partial fraction of

\frac{x ^{3} + x ^{2} + 8 x + 7}{6 - x ^{2} - x ^{4}} : \frac{9 - 10 2}{10 2 ( x + 2 )} - \frac{10 2 + 9}{10 2 ( x - 2 )} + \frac{5 x + 4}{5 ( x ^{2} + 3 )}

= \int \frac{9 - 10 2}{10 2 ( x + 2 )} - \frac{10 2 + 9}{10 2 ( x - 2 )} + \frac{5 x + 4}{5 ( 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{9 - 10 2}{10 2 ( x + 2 )} d x - \int \frac{10 2 + 9}{10 2 ( x - 2 )} d x + \int \frac{5 x + 4}{5 ( x ^{2} + 3 )} d x

\int \frac{9 - 10 2}{10 2 ( x + 2 )} d x = \frac{9 - 10 2}{10 2} ln x + 2

\int \frac{10 2 + 9}{10 2 ( x - 2 )} d x = \frac{10 2 + 9}{10 2} ln x - 2

\int \frac{5 x + 4}{5 ( x ^{2} + 3 )} d x = \frac{1}{2} ln x^{2} + 3 + \frac{4}{5 3} arctan (\frac{x}{3})

= \frac{9 - 10 2}{10 2} ln x + 2 - \frac{10 2 + 9}{10 2} ln x - 2 + \frac{1}{2} ln x^{2} + 3 + \frac{4}{5 3} arctan (\frac{x}{3})

Add a constant to the solution

= \frac{9 - 10 2}{10 2} ln x + 2 - \frac{10 2 + 9}{10 2} ln x - 2 + \frac{1}{2} ln x^{2} + 3 + \frac{4}{5 3} arctan (\frac{x}{3}) + C

\int (\frac{3}{10}) d t

\int \frac{11}{sec ( x ) - 1} d x

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

\int (1 + 3 x)^{4} d x

\int (- 2 x - 2) e^{x} d x

What is the integral of (x^3+x^2+8x+7)/(6-x^2-x^4) ?
The integral of (x^3+x^2+8x+7)/(6-x^2-x^4) is