What is the integral of x/(sqrt(x^2+4x-5)) ?

The integral of x/(sqrt(x^2+4x-5)) is sqrt(x^2+4x-5)-2ln(1/3 |sqrt(x^2+4x-5)+x+2|)+C

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integral of x/(sqrt(x^2+4x-5))

\int \frac{x}{x ^{2} + 4 x - 5} d x

x^{2} + 4 x - 5 - 2 ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + C

Solution steps

\int \frac{x}{x ^{2} + 4 x - 5} d x

Complete the square

x^{2} + 4 x - 5 : (x + 2)^{2} - 9

= \int \frac{x}{( x + 2 ) ^{2} - 9} d x

Apply u-substitution

: \int \frac{u - 2}{u ^{2} - 9} d u

= \int \frac{u - 2}{u ^{2} - 9} d u

Expand

\frac{u - 2}{u ^{2} - 9} : \frac{u}{u ^{2} - 9} - \frac{2}{u ^{2} - 9}

= \int \frac{u}{u ^{2} - 9} - \frac{2}{u ^{2} - 9} 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} - 9} d u - \int \frac{2}{u ^{2} - 9} d u

\int \frac{u}{u ^{2} - 9} d u = u^{2} - 9

\int \frac{2}{u ^{2} - 9} d u = 2 ln (\frac{1}{3} u^{2} - 9 + u)

= u^{2} - 9 - 2 ln (\frac{1}{3} u^{2} - 9 + u)

Substitute back

u = x + 2

= (x + 2)^{2} - 9 - 2 ln (\frac{1}{3} (x + 2)^{2} - 9 + x + 2)

Simplify

(x + 2)^{2} - 9 - 2 ln (\frac{1}{3} (x + 2)^{2} - 9 + x + 2) : x^{2} + 4 x - 5 - 2 ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2)

= x^{2} + 4 x - 5 - 2 ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2)

Add a constant to the solution

= x^{2} + 4 x - 5 - 2 ln (\frac{1}{3} x^{2} + 4 x - 5 + x + 2) + C

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What is the integral of x/(sqrt(x^2+4x-5)) ?
The integral of x/(sqrt(x^2+4x-5)) is sqrt(x^2+4x-5)-2ln(1/3 |sqrt(x^2+4x-5)+x+2|)+C