# How to Integrate Sqrt(1 + x^2) and Sqrt(a^2 + x^2)

TR Smith is a product designer and former teacher who uses math in her work every day.

The function y = sqrt(a^2 + x^2) is one half of a hyperbola (a two-piece conic section curve) oriented so that the line of symmetry is the y-axis. The integrals of the function y = sqrt(x^2 + 1) and that of the more general function y = sqrt(x^2 + a^2) have important applications in computing the arc lengths of curves, particularly parabolas. Though the function appears simple, the antiderivative of sqrt(x^2 + 1) is one of the trickiest to find, requiring both substitution and integration by parts.

Here we will work out step-by-step how to integrate sqrt(x^2 + 1). Once you know how to obtain the antiderivative of this function, you can use a simple substitution to extract the integral of the more general function sqrt(x^2 + a^2).

Graphs of y = sqrt(a^2 + x^2) for a=1 (red), a=2 (orange), a=3 (yellow), a=4 (green), a=5 (teal), a=6 (blue), a=7 (purple), a=8 (pink).

## Step 1

The first step in integrating sqrt(1 + x^2) is to make the trigonometric substitution x = tan(u) and dx = sec(u)^2 du. Then the integral of sqrt(x^2 + 1) becomes

∫ sqrt(x^2 + 1) dx

= ∫ sqrt(tan(u)^2 + 1) * sec(u)^2 du

= ∫ sqrt(sec(u)^2) * sec(u)^2 du

= ∫ sec(u)^3 du

This may look as intractable as before, but in the next step we will use integration by parts to find the antiderivative of secant cubed.

## Step 2

To integrate secant cubed, we use integration by parts. Recall the integration by parts formula,

∫ f * dg = fg - ∫ g * df

The task is to assign f, g, df, and df so that our integral becomes simpler. If we apply the assignment

f = sec(u)
dg = sec(u)^2 du
df = sec(u)tan(u) du
g = tan(u)

we can transform the integral of secant cubed into the equation

∫ sec(u)^3 du

= tan(u)sec(u) - ∫ sec(u)tan(u)^2 du

= tan(u)sec(u) - ∫ sec(u)[sec(u)^2 - 1] du

= tan(u)sec(u) - ∫ sec(u)^3 du + ∫ sec(u) du

= tan(u)sec(u) - ∫ sec(u)^3 du + Ln(sec(u) + tan(u))

This is an integral equation with the same integral term appearing on both sides. (Finding the integral of the secant function was explained in the linked article.) In the next step we will see how to resolve this equation.

## Step 3

The integral equation we obtained in the previous step is

∫ sec(u)^3 du = tan(u)sec(u) + Ln(sec(u) + tan(u)) - ∫ sec(u)^3 du

which contains the same integral on both sides. Thus, we can rewrite it as

2*∫ sec(u)^3 du = tan(u)sec(u) + Ln(sec(u) + tan(u))

Dividing both sides by 2 gives us the antiderivative of secant cubed:

∫ sec(u)^3 du = 0.5*tan(u)sec(u) + 0.5*Ln(sec(u) + tan(u)) + c

Now we just need to make the reverse substitution with u = arctan(x) to find the antiderivative of the original function, sqrt(1 + x^2). Since tan(arctan(x)) = x, we get

0.5*tan(u)sec(u) + 0.5*Ln(sec(u) + tan(u)) + c

= 0.5*x*sec(arctan(x)) + 0.5*Ln[sec(arctan(x)) + x) + c

This can be simplified even further using the identity sec(arctan(x)) = sqrt(x^2 + 1). This gives us

0.5*x*sqrt(x^2 + 1) + 0.5*Ln(sqrt(x^2 + 1) + x) + c

The above equation is the integral of the function sqrt(x^2 + 1).

## Integrating sqrt(a^2 + x^2)

The steps above give you the integral of the function sqrt(x^2 + a^2) when a = 1. To find the general form of the integral for an arbitrary value of a, we make the substitution x = au and dx = a*du. This gives us the integral equation

∫ sqrt(a^2 + x^2) dx

= ∫ sqrt(a^2 + (au)^2) * a * du

= ∫ a * sqrt(1 + u^2) * a * du

= (a^2) * ∫ sqrt(1 + u^2) du

= 0.5*a^2*[ u*sqrt(1 + u^2) + Ln(u + sqrt(1 + u^2)) ] + c

Now making the reverse substitution u = x/a, we get

0.5*a*x*sqrt(1 + (x/a)^2) + 0.5*a^2*Ln(x/a + sqrt(1 + (x/a)^2)) + c

With a little algebra you can express this in an alternate form that is more compact:

[x*sqrt(x^2 + a^2) + a^2*Ln(x + sqrt(x^2 + a^2))]/2 + c

## Example: Parabolic Arc Length

Let's find the arc length of the parabola y = (x/2)^2 from x = 0 to x = 16. The first step is the use the arc length integral formula

arc length = ∫ sqrt(1 + (y')^2) dx

In this example, we have y' = x/2. Thus, our integral is

∫ sqrt(1 + (x/2)^2) dx

= 0.5*∫ sqrt(4 + x^2) dx

= 0.5*∫ sqrt(2^2 + x^2) dx

Since we have a = 2, we use the general antiderivative formula for sqrt(a^2 + x^2) and plug in a = 2, which gives us

0.25x*sqrt(4+x^2) + Ln(x + sqrt(4+x^2))

Now we plug in the endpoints x = 16 and x = 0 and subtract to compute the arc length of the parabolic curve. This gives us

[0.25*16*sqrt(260) + Ln(16+sqrt(260))] - [0 + Ln(2)]
= 8*sqrt(65) + Ln(8 + sqrt(65))
≈ 67.274534.

Since parabolas are symmetric, the arc length from x = -16 to x = 16 is twice the number obtained above.

Graph of y = (x/2)^2 = 0.25x^2

## Example: How to Integrate sqrt(x^2 + x + 1)

The quadratic polynomial x^2 + x + 1 does not have the form z^2 + a^2, however, with a simple linear substitution we can put it into the correct form. Let x = u - 1/2 and du = dx. The integral becomes

∫ sqrt(x^2 + x + 1) dx
= ∫ sqrt(u^2 - u + 1/4 + u - 1/2 + 1) du
= ∫ sqrt(u^2 + 3/4)
= u*sqrt(u^2 + 3/4) + (3/4)Ln[ u + sqrt(u^2 + 3/4) ] + C

Now with the reverse substitution u = x + 1/2 we get

∫ sqrt(x^2 + x + 1) dx =
(x + 1/2)sqrt(x^2 + x + 1) + (3/4)Ln[ x + 1/2 + sqrt(x^2 + x + 1) ] + C

## Related Integral: Antiderivative of sqrt(1 + n/x)

Graph of y = sqrt(1 + 2/x)

Knowing the antiderivative of f(x) = sqrt(a^2 + x^2) lets us integrate the related function g(x) = sqrt(1 + n/x), where n is a positive number. Let's start with the substitution x = u^2 and dx = 2u du. This gives us the integral

∫ sqrt(1 + n/x) dx
= ∫ sqrt(1 + n/u^2) 2u du
= ∫ sqrt( 4u^2 + 4n) du
= 2 * ∫ sqrt(u^2 + n) du
= u*sqrt(u^2 + n) + n*Ln[u + sqrt(u^2 + n)] + C

Making the reverse substitution with u = sqrt(x) gives us

∫ sqrt(1 + n/x) dx =
sqrt(x^2 + nx) + n*Ln[sqrt(x) + sqrt(n + x)] + C

The graph of y = sqrt(1 + 2/x) is shown above.

3

0

2

4

## Popular

9

0

• ### Names of Geometric Shapes—With Pictures

30

0 of 8192 characters used