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

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

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

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

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

Solution steps

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

Factor

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

= \int \frac{x}{- ( - 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

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

Complete the square

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

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

Apply u-substitution

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

Expand

\frac{u - 1}{- u ^{2} + 2} : \frac{u}{- u ^{2} + 2} - \frac{1}{- u ^{2} + 2}

Apply the Sum Rule:

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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