What is the integral of (x^2+6x+4)/(x^4+8x^2+16) ?

The integral of (x^2+6x+4)/(x^4+8x^2+16) is 1/2 arctan(x/2)-3/(x^2+4)+C

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integral of (x^2+6x+4)/(x^4+8x^2+16)

\int \frac{x ^{2} + 6 x + 4}{x ^{4} + 8 x ^{2} + 16} d x

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

Solution steps

\int \frac{x ^{2} + 6 x + 4}{x ^{4} + 8 x ^{2} + 16} d x

Take the partial fraction of

\frac{x ^{2} + 6 x + 4}{x ^{4} + 8 x ^{2} + 16} : \frac{1}{x ^{2} + 4} + \frac{6 x}{( x ^{2} + 4 ) ^{2}}

= \int \frac{1}{x ^{2} + 4} + \frac{6 x}{( x ^{2} + 4 ) ^{2}} 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} + 4} d x + \int \frac{6 x}{( x ^{2} + 4 ) ^{2}} d x

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

\int \frac{6 x}{( x ^{2} + 4 ) ^{2}} d x = - \frac{3}{x ^{2} + 4}

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

Add a constant to the solution

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

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What is the integral of (x^2+6x+4)/(x^4+8x^2+16) ?
The integral of (x^2+6x+4)/(x^4+8x^2+16) is 1/2 arctan(x/2)-3/(x^2+4)+C