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

Updated on March 14, 2014
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TR Smith is a teacher and creator who uses mathematics in her line of work every day.

Integrals of the form 1/(a + b√x) are not as simple as those of the form 1/√(a + bx), though they look very similar. Rather than apply the power rule to sqrt(x) = x^(1/2), you need to use a combination of u-substitution and algebraic manipulation. It turns out the integrals of 1/(sqrt(x) + 1) and 1/(a + b*sqrt(x)) involve natural logarithms, which is not apparent from their deceptively simple forms.

If you know the antiderivative of 1/(1 + sqrt(x)) and that of 1/(a + b*sqrt(x)) you can compute the exact area under a curve, solve differential equations, and work out other calculus problems.


Graph of f(x) = 1/(sqrt(x) + 1)

Graph of the function f(x) = 1/(1 + sqrt(x))
Graph of the function f(x) = 1/(1 + sqrt(x))

Integral of 1/(1 + √x)

To integrate 1/(1 + sqrt(x)), we first apply the substitution x = u^2 and dx = 2u du. This gives us

1/(1 + sqrt(x)) dx
= 2u/(1+u) du

Now we need to use some algebra to put this in an equivalent form that's easier to integrate. Notice that

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

This gives us the equivalent integral expression

2u/(1+u) du
= 2 du - 2/(1+u) du
= 2u - 2*Ln|u + 1| + C
= 2*sqrt(x) - 2*Ln|1 + sqrt(x)| + C

Integral of 1/(a + b√x)

Integrating the more general function 1/sqrt(a + b*sqrt(x)) is done the same way as above. First we make the substitution x = u^2 and dx = 2u du. This gives us the new integral

2u/(a + bu) du.

Now we need to express this rational function in a different form.

2u/(a + bu)
= (2u + 2a/b - 2a/b)/(a + bu)
= (2u + 2a/b)/(bu + a) - (2a/b)/(a + bu)
= (2/b) - (2a/b)/(a + bu)

Integrating this form of the u-integral gives us

2u/(a + bu) du
= 2/b du - (2a/b)* 1/(a + bu) du
= (2/b)u - (2a/b^2)*Ln|a + bu| + c
= (2/b)*sqrt(x) - (2a/b^2)*Ln|a + b*sqrt(x)| + c

Example 1

Knowing the antiderivative of 1/(1 + sqrt(x)) allows us to completely solve the separable differential equation dy/dx = y/(1 + sqrt(x)). Using the technique of separation of variables we have

dy/dx = y/(1 + sqrt(x))

1/y dy = 1/(1 + sqrt(x)) dx

1/y dy = 1/(1 + sqrt(x)) dx

Ln(y) = 2*sqrt(x) - 2*Ln(1 + sqrt(x)) + c

y = e^[2*sqrt(x) - 2*Ln(1 + sqrt(x)) + c]

y = Ke^(2*sqrt(x))/(1 + 2*sqrt(x) + x)

Example 2

Let's find the exact area under the curve y = 1/(4 - sqrt(x)) from x = 0 to x = 9, shaded in yellow in the graph below.

This is a curve of the form y = 1/(a + b*sqrt(x)) where a = and b = -1. The antiderivative of this function is

-2*sqrt(x) - 8*Ln|4 - sqrt(x)|

Plugging the endpoints x = 9 and x = 0 into the expression above and subtracting gives us the area under the curve.

[-2*sqrt(9) - 8*Ln|4 - sqrt(9)|] - [-2*sqrt(0) - 8*Ln|4 - sqrt(0)|]
= [-6 - 8*Ln(1)] - [0 - 8*Ln(4)]
= 8*Ln(4) - 6
≈ 5.090355


Similar Integrals

Example 3

We can use the antiderivative of 1/(sqrt(x) + 1) to find the antiderivative of sqrt(x)/(x - 1). At first these functions may not look very similar, but with a little algebra we can transform it into a recognizable form. We use the fact that [sqrt(x) + 1]*[sqrt(x) - 1] = x - 1. This gives us

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

Now we get the integral

∫ sqrt(x)/(x - 1) dx
= ∫ 1/(1 + sqrt(x)) dx + ∫ 1/(x - 1) dx
= 2*sqrt(x) - 2*Ln|1 + sqrt(x)| + Ln|x - 1| + c

Making the change of variables w = x - 1, w + 1 = x, and dw = dx in the integral above gives us yet another antiderivative expression

∫ sqrt(w + 1)/w dw = 2*sqrt(w+1) - 2*Ln|1 + sqrt(w+1)| + Ln|w| + c

Thus, we can get a lot of mileage out of one integral.

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    • sehrm profile image

      sehrm 2 years ago from Los Angeles

      Can you solve this one for me, thanks!

      ∫ [(sqrt(x)+a)/(e^x-x)] dx

    • calculus-geometry profile image
      Author

      TR Smith 2 years ago from Germany

      Ha ha, non-integrable, unfortunately (or fortunately, depending on your viewpoint.)

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