Find the antiderivative of y= square root(1+square rootx).

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neela | High School Teacher | (Level 3) Valedictorian

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To find the antiderivative  of  sqrt(1+sqrtx)..

We put sqrt(1+sqrt x )  = y.

 (1+sqrt y) = y^2.

 sqrty = y^2-1.

Also differentiating sqrt(1+sqrx) = y, we get:

(1/2) (1/ sqrt(1+sqrt x) (1/2)(1/sqrtx) dx.  = dy.

Therefore dx = 4*sqrt(1+sqrtx) (sqrtx) dy

dx = 4y(y^2-1) dy.

Therefore Int sqrt(1+sqrt x) dx = Int y *4y(y^2-1) dy = 4Int  (y^4-y^3) dy

=4{ (y^5)/5 - y^4/4}+C

= 4y^4{y/5 - 1/4} + C

= (1/5)(1+sqrtx)^2 {4sqrt(1+sqrtx) -5} + C

Int sqrt(1+sqrtx) dx = (1/5)(1+sqrtx)^2{4 sqrt(1+sqrtx)-5} + C

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To find the antiderivative of the given function, we'll have to determine the indefinite integral.

Int  sqrt(1+sqrt x) dx

We'll substitute 1 + sqrt x = t

We'll differentiate both sides:

dx/2sqrt x = dt

dx = 2 sqrt x*dt

But sqrt x = t - 1

dx = 2(t - 1)dt

We'll re-write the integral in t:

Int sqrt t*2(t - 1)dt = 2Int sqrt t^3 dt - 2 Int sqrt t dt

 Int sqrt t*2(t - 1)dt = 2*t^(3/2 + 1)/(3/2 + 1) - 2*t^(1/2 + 1)/(1/2 + 1) + C

 Int sqrt t*2(t - 1)dt = 4t^(5/2)/5 - 4t^(3/2)/3 + C

 Int sqrt t*2(t - 1)dt = 4t^(3/2)( t/5 - 1/3) + C

Int sqrt(1+sqrt x)dx = 4(1 + sqrt x)^(3/2)[(1 + sqrt x)/5 - 1/3] + C

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