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

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

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

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

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

Solution steps

\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

= \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}

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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