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

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

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integral of 4/(x^2+2x)

\int \frac{4}{x ^{2} + 2 x} d x

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

Solution steps

\int \frac{4}{x ^{2} + 2 x} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} + 2 x} d x

Complete the square

x^{2} + 2 x : (x + 1)^{2} - 1

= 4 \cdot \int \frac{1}{( x + 1 ) ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 1} d u

= 4 \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x ln (z^{5} + x^{7}))

x y y^{^{'}} + 4 x^{2} + y^{2} = 0

y^{^{''}} + 6 y^{^{'}} + 10 = 0

\frac{d y}{d x} = \frac{3 x + 2 y}{3 x + 2 y + 2}

\int 6 x^{2} + 8 x^{3} + C d x

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