SUBSTITUTION AND ANTIDERIVATIVESI have to determine the antiderivative of the function f(x)=square root x/(1+square root x)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to add and subtract 1 to numerator, such that:

`int (sqrt x)/(1 + sqrt x)dx = int (sqrt x + 1 - 1)/(1 + sqrt x)dx`

Using the property of linearity of integral yields:

`int (sqrt x + 1 - 1)/(1 + sqrt x)dx = int (sqrt x + 1)/(1 + sqrt x)dx - int (dx)/(1 + sqrt x)dx`

Rationalizing the denominator `1 + sqrt x` yields:

`int (sqrt x + 1 - 1)/(1 + sqrt x)dx = int dx - int (1 - sqrt x)/(1 - x)dx`

`int (sqrt x + 1 - 1)/(1 + sqrt x)dx = int dx - int(dx)/(1 - x) - int (sqrt x)/(1 - x)dx`

You should come up with the following substitution to evaluate the integral ` int (sqrt x)/(1 - x)dx` such that:

`sqrt x = u => 1/(2sqrt x)dx = du => dx = 2u du`

`x = u^2 => -x = -u^2 => 1 - x = 1 - u^2`

`int (2u^2)/(1 - u^2)du = -2 int (1 - u^2 - 1)/(1 - u^2)du`

`int (2u^2)/(1 - u^2)du = -2 int (1 - u^2)/(1 - u^2)du + 2int (du)/(1 - u^2)`

`int (2u^2)/(1 - u^2)du = -u + 2ln|(1-u)/(1+u)| + c`

Substituting back `sqrt x` for u yields:

`int (sqrt x)/(1 - x)dx = -sqrt x + 2ln|(1 - sqrt x)/(1 + sqrt x)| + c`

`int (sqrt x)/(1 + sqrt x)dx = x - ln|1 - x| + sqrt x - 2ln|(1 - sqrt x)/(1 + sqrt x)| + c`

Hence, evaluating the given indefinite integral yields `int (sqrt x)/(1 + sqrt x)dx = x - ln|1 - x| + sqrt x - 2ln|(1 - sqrt x)/(1 + sqrt x)| + c.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll find the antiderivative of the given function evaluating the indefinite integral of this function.

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

We'll solve the integral using substitution technique. We'll substitute 1 + sqrt x = t.

We'll differentiate both sides:

dx/2sqrt x = dt

dx = (2sqrt x)dt

But sqrt x = t - 1

dx = 2(t - 1)dt

We'll re-write the integral;

Int f(x)dx = Int 2(t - 1)^2dt/t

We'll expand the square;

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

We'll apply the property of integrals to be additive:

2 Int (t^2 - 2t + 1)/t = 2Int t^2dt/t - 4Int tdt/t + 2Int dt/t

2 Int (t^2 - 2t + 1)/t = 2Int tdt - 4 Int dt + 2Int dt/t

2 Int (t^2 - 2t + 1)/t = 2*t^2/2 - 4t + 2ln|t| + C

We'll simplify and we'll get:

2 Int (t^2 - 2t + 1)/t = t^2 - 4t + 2ln|t| + C

Int f(x)dx = (1 + sqrt x)^2 - 4(1 + sqrt x) + ln (1 + sqrt x)^2 + C

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