Question

solve this problem as fast as possible
with detailed explanation
no spam pls
hurry up...........​

Answer 1

Answer:

(D) None  ( it is actually ln(α+1) ).

Step-by-step explanation:

Easy approach --- eliminating incorrect options

Call the given integral J(α).  Notice right away that when α=0, the integrand is identically zero, so J(0) = 0.  This rules out options (A) and (C).

Now consider what happens when α=1.  The integrand is undefined at x=1, taking the form 0/0, but we can use l'Hospital's rule to check the limit there.

As x --> 1,  (x-1) / (ln x)  -->  1 / (1/x) = x  --> 1.

Also, as x-->0,  x-1 --> -1 and ln(x) --> -∞,  so the integrand tends to 0.

Furthermore, the integrand is always positive on the interval (0,1).

So the region whose area is J(1) is contained within the unit square from (0,0) to (1,1).  It follows that J(1) < 1.

But 2ln( 1+1 ) = 2ln2 = ln4 > 1, since 4 > e.  This rules out option (B).

It follows that the answer is option (D).

Harder approach --- working out what the answer actually is

We can work out J(α) by making use of the fact that this is a function of α.  Using Leibniz's integral rule for passing differentiation under an integral sign, we can get the derivative J'(α).  Then we can work back to J(α) from there.

First, the derivative J'(α):

$\displaystyle J'(\alpha)=\frac{d}{d\alpha}J(\alpha) = \frac{d}{d\alpha}\int_0^1\frac{x^\alpha-1}{\ln x}\,dx\\\\{}\qquad=\int_0^1\frac{\partial}{\partial\alpha}\frac{x^\alpha-1}{\ln x}\,dx\\\\{}\qquad=\int_0^1\frac{(\ln x)\,x^\alpha}{\ln x}\,dx\\\\{}\qquad=\int_0^1 x^\alpha\,dx\\\\{}\qquad=\frac1{\alpha+1}$

From here, we have J(α) = ln(α+1) + C.

Now, when α=0, the integrand is identically zero, so J(0) = 0.  Thus C=0 and it follows that

J(α) = ln(α+1).