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

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

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

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

ln ∣ x ∣ - ln ∣ x + 6 ∣ + C

Solution steps

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

Take the constant out:

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

= 6 \cdot \int \frac{1 + x}{x ( x ^{2} + 7 x + 6 )} d x

Take the partial fraction of

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

= 6 \cdot \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

= 6 (\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 ∣

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

Simplify

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

= ln ∣ x ∣ - ln ∣ x + 6 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 6 ∣ + C

\frac{\partial}{\partial x} (cos^{4} (x^{3} y^{5}))

x \to - 8 lim (x^{2} - 3)

f (x) = 4 x^{3} - 8 x

a e^{b t}

f (x) = \frac{3}{x + 2}, at x = 3

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