Hi, I needed help with this improper integral.
lower limit : 2
upper limit : infinity
integral of (lnx)/(x+1) dx
Thank You so much :D
I saw this question and have not been able to get a handle on it
we know that ln(x) has an expansion formula
http://upload.wikimedia.org/math/6/d/b/6dbf56b2e5f34d298c1e71ddd93b3448.png
but it only good for |x-1| < 1
so we could create a series of terms
= 1/(x+1) ( (x-1) - (x-1)^2 /2 + (x-1)^3 / 3 - .... )
but to integrate each term would be a mess
- even Wolfram, usually very dependable, could not handle it.
http://integrals.wolfram.com/index.jsp?expr=ln%28x%29%2F%28x%2B1%29&random=false
To solve the improper integral ∫(lnx)/(x+1) dx with the lower limit of 2 and upper limit of infinity, you can apply the method of integration by parts.
To begin, let's rewrite the given integral using the substitution method. Let u = lnx and dv = dx/(x+1). Using the properties of logarithms, we can also replace dx in terms of du.
Therefore, du = (1/x) dx, and dv = dx/(x+1).
Now, let's differentiate u to find du and integrate dv to find v.
du = (1/x) dx,
Integrating dv = ∫dx/(x+1),
To integrate ∫dx/(x+1), we can apply the natural logarithm rule.
∫dx/(x+1) = ln|x+1|.
So, our equation becomes:
∫(lnx)/(x+1) dx = ∫u dv = uv - ∫v du,
∫(lnx)/(x+1) dx = lnx*ln|x+1| - ∫ln|x+1|*(1/x) dx.
Now let's evaluate the integral with the given limits.
∫(lnx)/(x+1) dx = [lnx*ln|x+1| - ∫ln|x+1|*(1/x) dx] from 2 to infinity.
To evaluate the integral at infinity, we need to determine the limit of the expression as x approaches infinity.
As x approaches infinity, lnx tends to infinity, and ln|x+1| also tends to infinity. So, we have an indeterminate form.
To solve this, we can apply L'Hopital's Rule. Differentiate the numerator and the denominator separately.
Taking the derivative of lnx in the numerator, we have (1/x).
Taking the derivative of ln|x+1| in the denominator, we have 1/(x+1).
Taking the limit as x approaches infinity again, we can apply L'Hopital's Rule once more.
Differentiating the numerator and the denominator once again, we get:
(1/x) / (1/(x+1)) = (x+1)/x = 1 + (1/x).
As x approaches infinity, this limit equals 1.
So, the integral evaluated at infinity is [lnx*ln|x+1| - ∫ln|x+1|*(1/x) dx] as x approaches infinity = (lninfinity * ln|infinity+1|) - 1.
Now, let's evaluate the integral at the lower limit of 2.
∫(lnx)/(x+1) dx = [lnx*ln|x+1| - ∫ln|x+1|*(1/x) dx] from 2 to infinity.
Substituting 2 into the equation, we have:
ln2 * ln|2+1| - ∫ln|2+1|*(1/2) dx.
Simplifying further, we have:
ln2 * ln3 - (ln3/2).
Now, we can subtract the value of the lower limit from the value at infinity to obtain the final result:
∫(lnx)/(x+1) dx = (lninfinity * ln|infinity+1|) - 1 - [ln2 * ln3 - (ln3/2)].