# How to Integrate ln(x)/sqrt(x)

The function f(x) = ln(x)/sqrt(x) can be alternatively written as

- f(x) = ln(x) * x^(-1/2)
- f(x) = ln(x) / x^(1/2)
- f(x) = ln(x) / √x
- f(x) = √x * ln(x) / x

These are all equivalent ways of writing the natural logarithm of x divided by the square root of x. While the derivative of this function can be easily found using the quotient rule, the antiderivative or integral is a little more complicated. This tutorial shows several to integrate ln(x)/sqrt(x) so you can evaluate definite integrals.

## Method 1: Substitution

Integrating with u-substitution is probably the easisest way to work out the integral of ln(x)/sqrt(x). Because the complicating factor in this integral is the square root, the natural first step is to make the substitution x = u^2 and dx = 2u du. This gives you

∫ ln(x)/sqrt(x) dx

= ∫ [ln(u^2)/sqrt(u^2)] * 2u du

= ∫ [2 * ln(u)/u] * 2u du

= ∫ 4 * ln(u) du

The antiderivative of the natural logarithm function is a classic integral that you can look up in any table of integrals (or work it out for yourself following this tutorial). For brevity we will skip the steps of integrating ln(u) and go directly to the final steps.

∫ 4 * ln(u) du

= 4u * ln(u) - 4u + C

Applying the reverse substitution gives us

4u * ln(u) - 4u + C

= 4*sqrt(x)*ln(sqrt(x)) - 4*sqrt(x) + C

= 2*sqrt(x)*ln(x) - 4*sqrt(x) + C

## Method 2: Integration by Parts

Some calculus students have trouble applying u-substitution to find antiderivatives, but are comfortable with the technique of integration by parts. To work out the antiderivative of ln(x)/sqrt(x) by parts, we can assign

u = ln(x)

dv = 1/sqrt(x) dx

which gives us

du = 1/x dx

v = 2*sqrt(x)

Plugging these into the formula ∫ u dv = uv - ∫ v du, we get

∫ ln(x)/sqrt(x) dx

= 2*ln(x)*sqrt(x) - 2*∫ 1/sqrt(x) dx

= 2*ln(x)*sqrt(x) - 4*sqrt(x) + C

This answer matches what we obtained using substitution.

## Method 3: Integration by Parts, Different Assignment

In the previous section we used integration by parts with the assignment u = ln(x) and dv = 1/sqrt(x) dx, which gave us the antiderivative in a few short and easy steps. But what would happen if you assigned the u and dv parts differently? Would you still get the same answer? Let's try this out with u = 1/sqrt(x) and dv = ln(x). This gives us

du = (-1/2)x^(-3/2) dx

v = x*ln(x) - x

Already this seems like a bad idea because these expressions are more complicated. Plugging all the pieces into the integration by parts formula gives us

∫ ln(x)/sqrt(x) dx

= sqrt(x)*ln(x) - sqrt(x) + (1/2)*∫ ln(x)/sqrt(x) dx - (1/2)*∫ 1/sqrt(x) dx

= sqrt(x)*ln(x) - 2*sqrt(x) + (1/2)*∫ ln(x)/sqrt(x) dx

As you can see, we have the integral in question on both sides of the equation, but with different coefficients. If we combine like terms, we get

(1/2)*∫ ln(x)/sqrt(x) dx = sqrt(x)*ln(x) - 2*sqrt(x)

Multiplying both sides by 2 gives us an answer consistent with the answer obtained in the previous sections. Even if you do integration by parts with the "wrong" assignment -- the harder assignment -- you can still extract the antiderivative.

## Example Integral Computation

Find the area under the curve y = ln(x)/sqrt(x) from x = 1 to x = m, where m is the global maximum. The first step in solving this problem is to figure out what m is. To find the global maximum of y = ln(x)/sqrt(x), we must take its derivative, set the expression equal to zero, and then solve for x. Doing so gives us

y' = x^(-3/2) - (1/2)ln(x)*x^(-3/2)

= [ 2 - ln(x) ] / [ 2*x^(3/2) ]

0 = [ 2 - ln(x) ] / [ 2*x^(3/2) ]

0 = 2 - ln(x)

2 = ln(x)

x = e^2

Therefore, m = e^2. Now we use the antiderivative of y = ln(x)/sqrt(x) to find the definite integral from x = 1 to x = e^2. Working this out gives us

2sqrt(e^2)ln(e^2) - 4sqrt(e^2) - 2sqrt(1)ln(1) + 4sqrt(1)

= 2*e*2 - 4*e - 0 + 4

= 4

The area under the curve y = ln(x)/sqrt(x) from x = 1 to x = e^2 is exactly 4.

## Integral of ln(x)/sqrt(x) from x = 0 to x = 1

The integral of ln(x)/sqrt(x) over the integral (0,1] is an improper integral because ln(x)/sqrt(x) is undefined at x = 0. However, the area between the curve and the x-axis over this interval is finite. It turns out that the area is exactly equal to -4; the negative sign is simply because the region lies below the x-axis. In absolute terms, the exact area is 4.

## Related Integrals

Finding the antiderivative of ln(x)/sqrt(x) is a special case of the more general problem of finding the integral of (x^a)ln(x), a power function of x multiplied by the natural logarithm of x, which is worked out in another tutorial. This article covers the particular case of a = -1/2.

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