What is the integral of x/(x^4+12x^3+36x^2) ?

The integral of x/(x^4+12x^3+36x^2) is 1/36 ln|x|-1/36 ln|x+6|+1/(6(x+6))+C

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integral of x/(x^4+12x^3+36x^2)

\int \frac{x}{x ^{4} + 12 x ^{3} + 36 x ^{2}} d x

\frac{1}{36} ln ∣ x ∣ - \frac{1}{36} ln ∣ x + 6 ∣ + \frac{1}{6 ( x + 6 )} + C

Solution steps

\int \frac{x}{x ^{4} + 12 x ^{3} + 36 x ^{2}} d x

Take the partial fraction of

\frac{x}{x ^{4} + 12 x ^{3} + 36 x ^{2}} : \frac{1}{36 x} - \frac{1}{36 ( x + 6 )} - \frac{1}{6 ( x + 6 ) ^{2}}

= \int \frac{1}{36 x} - \frac{1}{36 ( x + 6 )} - \frac{1}{6 ( x + 6 ) ^{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}{36 x} d x - \int \frac{1}{36 ( x + 6 )} d x - \int \frac{1}{6 ( x + 6 ) ^{2}} d x

\int \frac{1}{36 x} d x = \frac{1}{36} ln ∣ x ∣

\int \frac{1}{36 ( x + 6 )} d x = \frac{1}{36} ln ∣ x + 6 ∣

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

= \frac{1}{36} ln ∣ x ∣ - \frac{1}{36} ln ∣ x + 6 ∣ - (- \frac{1}{6 ( x + 6 )})

Simplify

= \frac{1}{36} ln ∣ x ∣ - \frac{1}{36} ln ∣ x + 6 ∣ + \frac{1}{6 ( x + 6 )}

Add a constant to the solution

= \frac{1}{36} ln ∣ x ∣ - \frac{1}{36} ln ∣ x + 6 ∣ + \frac{1}{6 ( x + 6 )} + C

\int_{- 2}^{2} (32 - 8 y^{2}) d y

\frac{d}{d x} ((\frac{3 x - 4}{5 x + 2})^{- 5})

f (x) = 5 x^{2} cos (x) + 4 x

\int_{1}^{- 1} 3 e^{x} d x

t^{2} y^{^{''}} + 2 t y^{^{'}} - 12 y = 0

What is the integral of x/(x^4+12x^3+36x^2) ?
The integral of x/(x^4+12x^3+36x^2) is 1/36 ln|x|-1/36 ln|x+6|+1/(6(x+6))+C