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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Factor

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

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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