Find the integral of square root of (x^2+4x-5)

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neela | High School Teacher | (Level 3) Valedictorian

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To find the integral of square root of (x^2+4x-5).

Int f(x)dx = Int ((x^2+4x+5) dx.

Int f(x) dx = Int sqrt{x+2)^2 -2^2-9} dx

Int f(x)dx = Int sqrt{ x+2)^2-3^2} dx.

Int f(x)dx = Int sqrt(t^2-a^2) dt, where t = x+2 , a = 3 and dx= dt.

Int f(x)dx = (1/2)t(t^2-a^2)^(1/2) - a^2/2cosh^(-2) (t/3) +C.

Int f(x) dx = (1/2)(x+2)sqrt{x^2+4x+5} - (9/2)cosh^(-2) [(x+2)/3] +C.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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