# Integral of a square root.Determine the integral of square root of (x^2+4x-5).

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We'll re-write the expression x^2+4x-5 as following:

x^2+4x-5 = (x + 2)^2 - 9

We'll re-write the integral:

Int sqrt[ (x + 2)^2 - 9 ]dx

We'll substitute x + 2 = t

We'll differentiate both sides:

dx = dt

We'll re-write the integral in t:

Int sqrt (t^2 - 9) dt

We'll solve the integral by parts:

Int udv = uv - Int vdu

We'll put u = sqrt(t^2 - 9)

We'll differentiate:

du = tdt/sqrt(t^2 - 9)

We'll put dv = dt

v = t

We'll substitute u,v,du,dv into the formula:

I = Int sqrt (t^2 - 9) dt = tsqrt(t^2 - 9) - Int t*tdt/sqrt(t^2 - 9)

We'll note Int t*tdt/sqrt(t^2 - 9) = I1

I1 = Int (t^2 - 9 + 9)dt/sqrt(t^2 - 9)

I1 = I - 9Int dt/sqrt(t^2 - 9)

I = tsqrt(t^2 - 9) - I - 9ln [t+sqrt(t^2-9)]

We'll add I both sides:

2I = tsqrt(t^2 - 9) - 9ln [t+sqrt(t^2-9)]

I = {tsqrt(t^2 - 9) - 9ln [t+sqrt(t^2-9)]}/2 + C

**I = {(x+2)sqrt[(x+2)^2 - 9] - 9ln [x+2+sqrt[(x+2)^2-9]}/2 + C**