What is the integral of x/(sqrt(x^2+2x-3)) ?

The integral of x/(sqrt(x^2+2x-3)) is sqrt(x^2+2x-3)-ln(1/2 |sqrt(x^2+2x-3)+x+1|)+C

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integral of x/(sqrt(x^2+2x-3))

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

x^{2} + 2 x - 3 - ln (\frac{1}{2} x^{2} + 2 x - 3 + x + 1) + C

Solution steps

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

Complete the square

x^{2} + 2 x - 3 : (x + 1)^{2} - 4

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u ^{2} - 4} : \frac{u}{u ^{2} - 4} - \frac{1}{u ^{2} - 4}

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

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

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

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

Substitute back

u = x + 1

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

Simplify

(x + 1)^{2} - 4 - ln (\frac{1}{2} (x + 1)^{2} - 4 + x + 1) : x^{2} + 2 x - 3 - ln (\frac{1}{2} x^{2} + 2 x - 3 + x + 1)

= x^{2} + 2 x - 3 - ln (\frac{1}{2} x^{2} + 2 x - 3 + x + 1)

Add a constant to the solution

= x^{2} + 2 x - 3 - ln (\frac{1}{2} x^{2} + 2 x - 3 + x + 1) + C

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What is the integral of x/(sqrt(x^2+2x-3)) ?
The integral of x/(sqrt(x^2+2x-3)) is sqrt(x^2+2x-3)-ln(1/2 |sqrt(x^2+2x-3)+x+1|)+C