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How to Integrate x*sqrt(x+1) and x/sqrt(x+1)

Updated on October 08, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

How to Integrate Any Function of the Form x*sqrt(ax+b) and x/sqrt(ax+b)

The integrals of x*sqrt(x+1) and x/sqrt(x+1) are easy to work out with a simple linear substitution. These functions are particular examples of functions of the form x*sqrt(ax+b) and x/sqrt(ax+b), where 'a' and 'b' are some constants. While there is no way to simplify these functions algebraically in their original form, they become very easy to integrate after you make a change of variable. Here we show how to integrate x*sqrt(x+1) and x/sqrt(x+1) step by step, as well as the more general functions x*sqrt(ax+b) and x/sqrt(ax+b). With the general formula for the antiderivative, you can compute exact areas under curves.

Antiderivative of x*sqrt(x+1)

The first step is to apply the linear u-substitution u = x+1 and du = dx. This gives us the integral transformation

∫ x*sqrt(x+1) dx
= ∫ (u-1)sqrt(u) du
= ∫ [u*sqrt(u) - sqrt(u)] du
= ∫ [u^(3/2) - u^(1/2)] du

The antiderivative of u^(3/2) - u^(1/2) is (2/5)u^(5/2) - (2/3)u^(3/2) + C. Making the reverse substitution gives us

(2/5)u^(5/2) - (2/3)u^(3/2) + C
= (2/5)(x+1)^(5/2) - (2/3)(x+1)^(3/2) + C
= (2/15)(3x^2 + x - 2)sqrt(x+1) + C

Antiderivative of x/sqrt(x+1)

As in the previous problem, we can make the u-substitution u = x+1, x = u-1, dx = du. This gives us an easy power function integral.

∫ x/sqrt(x+1) dx
= ∫ (u-1)/sqrt(u) du
= ∫ [sqrt(u) - 1/sqrt(u)] du
= ∫ [u^(1/2) - u^(-1/2)] du
= (2/3)u^(3/2) - 2u^(1/2) + C
= (2/3)(x+1)^(3/2) - 2(x+1)^(1/2) + C
= (2/3)(x-2)sqrt(x+1) + C


How to Integrate x*sqrt(ax+b) and x/sqrt(ax+b)

Following the method above, to integrate the function f(x) = x*sqrt(ax+b), you make the substitution u = ax+b, x = (u-b)/a, and dx = (1/a)du. This gives you

∫ x*sqrt(ax+b) dx
= (1/a^2) * ∫ (u-b)*sqrt(u) du
= (1/a^2)*[(2/5)u^(5/2) - (2b/3)u^(3/2)] + C
= (1/a^2)*[(2/5)(ax+b)^(5/2) - (2b/3)(ax+b)^(3/2)] + C
= [2/(15a^2)] * (3ax-2b) * (ax+b)^(3/2) + C

∫ x/sqrt(ax+b) dx
= (1/a^2) * ∫ (u-b)/sqrt(u) du
= (1/a^2)*[(2/3)u^(3/2) - (2b)u^(1/2)] + C
= (1/a^2)*[(2/3)(ax+b)^(3/2) - (2b)(ax+b)^(1/2)] + C
= [2/(3a^2)] * (ax-2b) * (ax+b)^(1/2) + C

Example: Integral of x*sqrt(1-x)

Find the area under the curve x*sqrt(1-x) from x = 0 to x = 1. The area is shown in the graph below.

Since this is an integral of the form x*sqrt(ax+b) with b = 1 and a = -1, we can use the formula

∫ x*(ax+b) dx
= [2/(15a^2)] * (3ax - 2b) * (ax + b)^(3/2) + C

and plug in the values a = -1 and b = 1. This gives us the specific antiderivative

∫ x*sqrt(1 - x) dx
= (2/15) * (-3x - 2) * (1 - x)^(3/2) + C

Now we plug in the end points x = 1 and x = 0 into the antiderivative expression and subtract to obtain the exact area under the curve from 0 to 1. This gives us

(2/15)(-3 - 2)(1-1)^(3/2) - (2/15)(0 - 2)(1-0)^(3/2) = 4/15


Example: Integral of x/sqrt(2x+1)

Find the area above the curve x/sqrt(2x+1) and below the x-axis from between x = -1/2 and x = 0. As shown in the graph below, this region is unbounded since the function y = x/sqrt(2x+1) has a vertical asymptote at x = -1/2, but the area of the region is finite.

Using the formula for antiderivatives of functions of the form x/sqrt(ax+b), with a = 2 and b = 1, we have

∫ x/sqrt(ax + b) dx
= [2/(3a^2)] * (ax - 2b) * (ax + b)^(1/2) + C

and specifically

∫ x/sqrt(2x + 1) dx
= (1/6) * (2x - 2) * (2x + 1)^(1/2) + C

Plugging in the endpoints x = 0 and x = -1/2 and subtracting gives us

(1/6)(0 - 2)(0 + 1)^(1/2) - (1/6)(-1 - 2)(-1 + 1)^(1/2) = -1/3

The area in this example is negative because the region lies below the x-axis.

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