What is the indefinite integral of f(x)=1/(1+squareroot(x-1))?

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given f(x) = 1/(1+sqrt(x-1))

We need to find  the indefininte integral of f(x).

==> intg f(x) = intg (1/(1+sqrt(x-1)  dx

Let us assume that u= sqrt(x-1) ==> du = 1/sqrt(x-1) dx = 1/2u dx ==> 2u du = dx

Let us substitue:

==> intg f(x) = intg 1/(1+u) * 2u du

                     = ing 2u/(1+u)  du

                      = 2 intg u/(1+u)  du

                       = 2 intg (u+1 -1)/(1+u)  du

                      = 2[ intg (u+1)/(1+u) du - intg (1/1+u)  du]

                      = 2 [ intg du - intg (1/(1+u) du

                      = 2[ u - ln (1+u) ] + C

                         = 2u - 2ln (1+u) + C

Now we will substitute with u= sqrt(x-1)

==> intg f(x) = 2sqrt(x-1) - 2ln (sqrt(x-1) + 1) + C

 

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Int f(x) dx =Int dx/[1+sqrt(x-1)]

We'll substitute sqrt(x - 1) = t

We'll raise to square both sides:

x - 1 = t^2

x = t^2 + 1

We'll differentiate both sides:

dx = 2tdt

We'll re-write the integral in t:

Int 2tdt/(1+t) = 2Int tdt/(1+t)

We'll add and subtract 1 to numerator of the ratio:

2Int (t+1 - 1)dt/(1+t) = 2Int (t+1)dt/(1+t) - 2Int dt/(t+1)

We'll simplify and we'll get:

Int f(x) dx = 2Int dt - 2ln |t + 1|

We'll change the variable t:

Int f(x) dx = 2*sqrt(x - 1) - 2ln [sqrt(x - 1) + 1] + C

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