What is the integral of (6x^3)/(x^4+8x^2+16) ?

The integral of (6x^3)/(x^4+8x^2+16) is 6(1/32 x^3sin(2arctan(x/2))-1/8 x^2+1/2 ln|x^2+4|-1/2)+C

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integral of (6x^3)/(x^4+8x^2+16)

\int \frac{6 x ^{3}}{x ^{4} + 8 x ^{2} + 16} d x

6 (\frac{1}{32} x^{3} sin (2 arctan (\frac{x}{2})) - \frac{1}{8} x^{2} + \frac{1}{2} ln x^{2} + 4 - \frac{1}{2}) + C

Solution steps

\int \frac{6 x ^{3}}{x ^{4} + 8 x ^{2} + 16} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{x ^{3}}{x ^{4} + 8 x ^{2} + 16} d x

Apply Integration By Parts

: \frac{1}{16} x^{3} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) - \int \frac{3}{16} x^{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) d x

= 6 (\frac{1}{16} x^{3} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) - \int \frac{3}{16} x^{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) d x)

\int \frac{3}{16} x^{2} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) d x = \frac{1}{16} (x^{3} arctan (\frac{x}{2}) + 2 x^{2} + 8 - 8 ln x^{2} + 4)

= 6 (\frac{1}{16} x^{3} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) - \frac{1}{16} (x^{3} arctan (\frac{x}{2}) + 2 x^{2} + 8 - 8 ln x^{2} + 4))

Expand

\frac{1}{16} x^{3} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))) - \frac{1}{16} (x^{3} arctan (\frac{x}{2}) + 2 x^{2} + 8 - 8 ln x^{2} + 4) : \frac{1}{32} x^{3} sin (2 arctan (\frac{x}{2})) - \frac{1}{8} x^{2} + \frac{1}{2} ln x^{2} + 4 - \frac{1}{2}

= 6 (\frac{1}{32} x^{3} sin (2 arctan (\frac{x}{2})) - \frac{1}{8} x^{2} + \frac{1}{2} ln x^{2} + 4 - \frac{1}{2})

Add a constant to the solution

= 6 (\frac{1}{32} x^{3} sin (2 arctan (\frac{x}{2})) - \frac{1}{8} x^{2} + \frac{1}{2} ln x^{2} + 4 - \frac{1}{2}) + C

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What is the integral of (6x^3)/(x^4+8x^2+16) ?
The integral of (6x^3)/(x^4+8x^2+16) is 6(1/32 x^3sin(2arctan(x/2))-1/8 x^2+1/2 ln|x^2+4|-1/2)+C