Find the antiderivative of x^1/2/(x-1)

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hala718's profile pic

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

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f(x) = x^1/2 / (x-1)

We need to find F(x) such that:

F(x) = intg f(x) dx

==> F(x) = intg (x^1/2)/ (x-1)  dx

Let us rewrite:

==> (x-1) = (sqrtx - 1)(sqrtx + 1)

==> F(x) = intg (x^1/2)/(sqrtx-1)(sqrtx+1) dx

let t= sqrtx

==> dt= 1/2sqrtx dx

==> dt= 1/2t dx

==> dx= 2t dt :

==> F(x) = intg [(t)/(t-1)(t+1)] * 2t dt

                = intg 2t^2/(t-1)  * intg 1/(t+1)                               

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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F(x) is such a function such as dF/dx = f(x).

We'll calculate the function F(x) integrating the given function:

Int f(x)dx = F(x) + C

Int sqrtx dx/(x-1)

We'll write the denominator of the function as a difference of squares:

x - 1 = (sqrtx - 1)(sqrtx + 1)

We'll re-write the integral:

Int sqrtx dx/(sqrtx - 1)(sqrtx + 1)

We'll add and subtract 1 to the numerator:

Int (sqrtx + 1 - 1 )dx/(sqrtx - 1)(sqrtx + 1)

We'll re-group the terms in a convenient way:

Int (sqrtx + 1 - 1 )dx/(sqrtx - 1)(sqrtx + 1) = Int(sqrtx + 1)dx/(sqrtx - 1)(sqrtx + 1) - Int dx/(sqrtx - 1)(sqrtx + 1)

We'll simplify and we'll get:

Int (sqrtx + 1 - 1 )dx/(sqrtx - 1)(sqrtx + 1) = Int dx - Int dx/(x-1)

Int f(x)dx = x - ln|x-1| + C

F(x) = x - ln|x-1| + C

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the  antiderivative of x^(1/2)/(x-1).

The antiderivative of x^(1/2)/(x-1) is f(x) = Integral {x^(1/2) /(x-1) } dx.

Let  x ^(1/2) = t. Then x = t^2.

We differentiate x with respect to t.

(1/2) dx/(x^(/2) = dt.

dx = 2tdt.

The given integral under the transfotmation becomes:

Integral t*2tdt/(t^2-1) =  Integral {2t^2/t^2-1)}.

2Integral t^2 dt/(t^2-1)= 2Integral { t^2-1)+1/(t^2-1)}dt

= 2Integral 2dt  + Integral dt/(t^2-1).

= 2t + 2Int { 1/2(t-1) -1/2(t+1)}.

= 2t + log (t-1)/(t+1) .

Therefore antiderivative of x^(1/2)/(x-1) = 2x^(1/2) +log [(x^(1/2)-1]/[(x^(1/2)+1] +  C , where C is the  constant of integration.

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