# How to Integrate sqrt(x)/(x+1) and sqrt(x)/(x-1)

While it is easy to integrate functions like x*sqrt(x+1) and x/sqrt(x+1) with a simple linear substitution, it is not as straight forward to integrate the square root of x over (x+1), (x-1), or any other linear term of the form ax + b. However, with a different substitution and a knowledge of common integrals, you can find the antiderivative of sqrt(x)/(x+1) and sqrt(x)/(x-1), as well as any function of the form sqrt(x)/(ax+b). Here we work out the integration step by step and show some examples of how to use antiderivatives in calculus problems.

## Intregral of sqrt(x)/(x+1)

The first step in integrating is making the change of variables x = u^2, which gives us dx = 2u du. The original integral of sqrt(x)/(x+1) is transformed into

∫ sqrt(x)/(x+1) dx

= ∫ u/(u^2 + 1) * 2u du

= ∫ (2u^2)/(u^2 + 1) du

= ∫ (2u^2 + 2)/(u^2 + 1) du - ∫ 2/(u^2 + 1) du

= ∫ 2 du - 2*∫ 1/(u^2 + 1) du

= 2u - 2*arctan(u) + C

Now we make the back substitution with u = sqrt(x). This gives us

∫ sqrt(x)/(x+1) dx

= 2*sqrt(x) - 2*arctan(sqrt(x)) + C

## Integral of sqrt(x)/(x-1)

As with the previous integral, the easiest method is to start with the simple u-substitution x = u^2 and dx = 2u du. This gives us

∫ sqrt(x)/(x-1) dx

= ∫ u/(u^2 - 1) * 2u du

= ∫ (2u^2)/(u^2 - 1) du

= ∫ (2u^2 - 2)/(u^2 - 1) du + ∫ 2/(u^2 - 1) du

= ∫ 2 du + ∫ 2/(u^2 - 1) du

Since 2/(u^2 - 1) = 1/(u-1) - 1/(u+1), we can further breakdown the integral into

∫ 2 du + ∫ 1/(u-1) du - ∫ 1/(u+1) du

= 2u + Ln |u-1| - Ln |u+1| + C

= 2u + Ln |(u-1)/(u+1| + C

With the back substitution of u = sqrt(x), we have

∫ sqrt(x)/(u-1) dx

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

## How to Find the Antiderivatives of sqrt(x)/(ax+b) and sqrt(x)/(ax-b)

The square root of x divided by a linear polynomial ax + b or ax - b always has an antiderivative, but the form of the antiderivative depends on whether there is a plus or minus in the denominator. If a and b are positive numbers and the denominator is ax + b, then the integral of sqrt(x)/(ax+b) has an inverse tangent (arctangent) term. if and b are positive and the denominator is ax - b, the integral expression has a natural logarithm term. The general formulas are found by making the same substitution of x = u^2 and dx = 2u du. This gives you the integrals

∫ sqrt(x)/(ax + b) dx

= (2/a)sqrt(x) - sqrt(4b/a^3)*arctan[sqrt(ax/b)] + C

∫ sqrt(x)/(ax + b) dx

= (2/a)sqrt(x) + sqrt(b/a^3)*Ln |(sqrt(ax/b) - 1)/(sqrt(ax/b) + 1)| + C

## How to Use Antiderivatives to Find Areas Under Curves

Knowing the antiderivative of a function allows you to find the area between the curve and the x-axis over a given interval. For example, let's find the area under the curve sqrt(x)/(x+2) from between x = 0 and x = 4. This region is shown in the image below.

Using a = 1 and b = 2, and the first form of the antiderivative, we have

∫ sqrt(x)/(x+2) dx

= 2*sqrt(x) - sqrt(8)*arctan(sqrt(x/2))

Plugging in the endpoints x = 4 and x = 0 and subtracting gives you the exact area under the curve.

2*sqrt(4) - sqrt(8)*arctan(sqrt(4/2)) - 0 + 0

= 4 - 2*sqrt(2)*arctan(sqrt(2))

≈ 1.29795656

## Using Antiderivatives to Solve Differential Equations

Antiderivatives are useful in solving separable differential equations and differential equations of the form y' + f(x)*y = g(x), which uses an integrating factor. For example, let's use this technique to solve the differential equation

y' + y*sqrt(x)/(x-1) = 1

The integrating factor is e^[ ∫ sqrt(x)/(x-1) dx ]. Working out this expression gives us

e^[ ∫ sqrt(x)/(x-1) dx ]

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

= e^(2*sqrt(x)) + (sqrt(x) - 1)/(sqrt(x) + 1)

Multiplying both sides of the original differential equation gives us

{ y*[e^(2*sqrt(x)) + (sqrt(x) - 1)/(sqrt(x) + 1)] }*'*

= e^(2*sqrt(x)) + (sqrt(x) - 1)/(sqrt(x) + 1)

Integrating both sides gives us

y*[e^(2*sqrt(x)) + (sqrt(x) - 1)/(sqrt(x) + 1)]

= x - 4*sqrt(x) + sqrt(x)*e^(2*sqrt(x)) - 0.5e^(2*sqrt(x)) + 4*Ln(sqrt(x)+1) + C

Dividing both sides by [e^(2*sqrt(x)) + (sqrt(x) - 1)/(sqrt(x) + 1)] gives the general solution. As you can see, a simple-looking differential equation can have a very complicated expression as its solution.

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