What is the integral of 6/(x^2+10x+16) ?

The integral of 6/(x^2+10x+16) is ln|x+2|-ln|x+8|+C

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integral of 6/(x^2+10x+16)

\int \frac{6}{x ^{2} + 10 x + 16} d x

ln ∣ x + 2 ∣ - ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{6}{x ^{2} + 10 x + 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{1}{x ^{2} + 10 x + 16} d x

Take the partial fraction of

\frac{1}{x ^{2} + 10 x + 16} : \frac{1}{6 ( x + 2 )} - \frac{1}{6 ( x + 8 )}

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

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

\int \frac{1}{6 ( x + 8 )} d x = \frac{1}{6} ln ∣ x + 8 ∣

= 6 (\frac{1}{6} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x + 8 ∣)

Simplify

6 (\frac{1}{6} ln ∣ x + 2 ∣ - \frac{1}{6} ln ∣ x + 8 ∣) : ln ∣ x + 2 ∣ - ln ∣ x + 8 ∣

= ln ∣ x + 2 ∣ - ln ∣ x + 8 ∣

Add a constant to the solution

= ln ∣ x + 2 ∣ - ln ∣ x + 8 ∣ + C

\int 6 sin^{4} (x) d x

\frac{d}{d x} ((12 x^{2} - 12) (9 - 4 x))

\frac{\partial}{\partial x} (x^{2} - 2 x y + 3 y^{2} x^{3})

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

\frac{d y}{d x} + 12 y = 6

What is the integral of 6/(x^2+10x+16) ?
The integral of 6/(x^2+10x+16) is ln|x+2|-ln|x+8|+C