What is the integral of (7x+1)/(x^3+6x^2+12x+8) ?

The integral of (7x+1)/(x^3+6x^2+12x+8) is 7(-x/(2(x+2)^2)-1/(2(x+2)))-1/(2(x+2)^2)+C

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integral of (7x+1)/(x^3+6x^2+12x+8)

\int \frac{7 x + 1}{x ^{3} + 6 x ^{2} + 12 x + 8} d x

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

Solution steps

\int \frac{7 x + 1}{x ^{3} + 6 x ^{2} + 12 x + 8} d x

Expand

\frac{7 x + 1}{x ^{3} + 6 x ^{2} + 12 x + 8} : \frac{7 x}{x ^{3} + 6 x ^{2} + 12 x + 8} + \frac{1}{x ^{3} + 6 x ^{2} + 12 x + 8}

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{7 x}{x ^{3} + 6 x ^{2} + 12 x + 8} d x + \int \frac{1}{x ^{3} + 6 x ^{2} + 12 x + 8} d x

\int \frac{7 x}{x ^{3} + 6 x ^{2} + 12 x + 8} d x = 7 (- \frac{x}{2 ( x + 2 ) ^{2}} - \frac{1}{2 ( x + 2 )})

\int \frac{1}{x ^{3} + 6 x ^{2} + 12 x + 8} d x = - \frac{1}{2 ( x + 2 ) ^{2}}

= 7 (- \frac{x}{2 ( x + 2 ) ^{2}} - \frac{1}{2 ( x + 2 )}) - \frac{1}{2 ( x + 2 ) ^{2}}

Add a constant to the solution

= 7 (- \frac{x}{2 ( x + 2 ) ^{2}} - \frac{1}{2 ( x + 2 )}) - \frac{1}{2 ( x + 2 ) ^{2}} + C

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What is the integral of (7x+1)/(x^3+6x^2+12x+8) ?
The integral of (7x+1)/(x^3+6x^2+12x+8) is 7(-x/(2(x+2)^2)-1/(2(x+2)))-1/(2(x+2)^2)+C