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

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

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

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

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

Solution steps

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

Simplify

\frac{x + 1}{x ^{3} + 7 x ^{2} + 6 x} : \frac{1}{x ( x + 6 )}

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

Take the partial fraction of

\frac{1}{x ( x + 6 )} : \frac{1}{6 x} - \frac{1}{6 ( x + 6 )}

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

\int \frac{1}{6 x} d x = \frac{1}{6} ln ∣ x ∣

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{4}{5 x ^{6}})

f (x) = 2 - 6 x^{2}, (- 5, - 148)

\frac{d}{d x} ((3 x^{4} - 6 x)^{3})

y = 4 e^{x} cos (x)

y = \frac{ln ( x )}{x}

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