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

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

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

\int \frac{1}{2 x + x ^{2}} d x

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

Solution steps

\int \frac{1}{2 x + x ^{2}} d x

Complete the square

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

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

Apply u-substitution

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

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

Factor

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

= \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

= - \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}

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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

\int e^{3 x} x + e^{x} d x

\int \frac{x}{27 x ^{2} - 1} d x

x \to 0 + lim (- x + 3)

\frac{d ^{2} y}{d x ^{2}} + 25 y = 0

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