What is the integral of (2x)/(x^2+6x+5) ?

The integral of (2x)/(x^2+6x+5) is 2(5/4 ln|x+5|-1/4 ln|x+1|)+C

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integral of (2x)/(x^2+6x+5)

\int \frac{2 x}{x ^{2} + 6 x + 5} d x

2 (\frac{5}{4} ln ∣ x + 5 ∣ - \frac{1}{4} ln ∣ x + 1 ∣) + C

Solution steps

\int \frac{2 x}{x ^{2} + 6 x + 5} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{x ^{2} + 6 x + 5} d x

Complete the square

x^{2} + 6 x + 5 : (x + 3)^{2} - 4

= 2 \cdot \int \frac{x}{( x + 3 ) ^{2} - 4} d x

Apply u-substitution

: \int \frac{u - 3}{u ^{2} - 4} d u

= 2 \cdot \int \frac{u - 3}{u ^{2} - 4} d u

Take the partial fraction of

\frac{u - 3}{u ^{2} - 4} : \frac{5}{4 ( u + 2 )} - \frac{1}{4 ( u - 2 )}

= 2 \cdot \int \frac{5}{4 ( u + 2 )} - \frac{1}{4 ( u - 2 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{5}{4 ( u + 2 )} d u - \int \frac{1}{4 ( u - 2 )} d u)

\int \frac{5}{4 ( u + 2 )} d u = \frac{5}{4} ln ∣ u + 2 ∣

\int \frac{1}{4 ( u - 2 )} d u = \frac{1}{4} ln ∣ u - 2 ∣

= 2 (\frac{5}{4} ln ∣ u + 2 ∣ - \frac{1}{4} ln ∣ u - 2 ∣)

Substitute back

u = x + 3

= 2 (\frac{5}{4} ln ∣ x + 3 + 2 ∣ - \frac{1}{4} ln ∣ x + 3 - 2 ∣)

Simplify

2 (\frac{5}{4} ln ∣ x + 3 + 2 ∣ - \frac{1}{4} ln ∣ x + 3 - 2 ∣) : 2 (\frac{5}{4} ln ∣ x + 5 ∣ - \frac{1}{4} ln ∣ x + 1 ∣)

= 2 (\frac{5}{4} ln ∣ x + 5 ∣ - \frac{1}{4} ln ∣ x + 1 ∣)

Add a constant to the solution

= 2 (\frac{5}{4} ln ∣ x + 5 ∣ - \frac{1}{4} ln ∣ x + 1 ∣) + C

3 t cos (3 t)

x \to 1 lim (e^{x - 2})

\frac{\partial}{\partial x} (2 x^{2} y)

y^{^{''}} + 16 y = 4 sec (4 t)

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

What is the integral of (2x)/(x^2+6x+5) ?
The integral of (2x)/(x^2+6x+5) is 2(5/4 ln|x+5|-1/4 ln|x+1|)+C