# Sir can you tell the solution of this challenging question? The answer given is ln 8

KAPIL MANDAL
132 Points
6 years ago
∫{(x7+1) / ln (x) }dx
At first, you put ln(x)=z
This gives, x=ez and x-1dx=dz
Then, ∫{(x7+1) / ln (x) }dx =∫{(x8+x) / z}dz = ∫{(e8z+ez) / z}dz.
Now you seperate those.
∫{(e8z+ez) / z}dz =  ∫{(e8z) / z}dz  +   ∫{(ez) / z}dz.  ….(1)
The formula you need to know is,
∫ecx/x dx = ln (x) +∑(cx)n/(n.n!)
Forget about summation part. it is going to be cancelled when you take both the limits. So, for this particular problem,
∫ecx/x dx = ln (x) .
So, the second part of eqn (1) is going to be cancelled as ln (1) = 0
Now you have only the first part.
Now simple algebra will lead you to the result.
Any problem you face, write here.

28 Points
6 years ago
Sir if we apply limit 1 to 0 to ∫ecx/x dx = ln (x) +∑(cx)n/(n.n!) that is in ln(x) we will get ln1 - ln 0. . But the aanswer is given ln 8, and Sir how to derive the relation ∫ecx/x dx = ln (x) +∑(cx)n/(n.n!) ?
KAPIL MANDAL
132 Points
6 years ago
When you change the 8z part to some y(say), integral should come like ln(8x). I got the formula from wiki. I’m sending yu the page url. https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
mycroft holmes
272 Points
6 years ago
I think you are talking about

$\int_0^1 \frac{x^7-1}{\ln x} \ \ dx$
Please check your question. This above integral has the value of ln 8. If you get back on this I can help you with the proof
28 Points
6 years ago
Yes sir it is minus sign not plus............................................................,............
mycroft holmes
272 Points
6 years ago
This uses a technique called Differentiating under the integral sign.

Consider the generalized problem of finding the integral

$I(a) = \int_0^1 \frac{x^a-1}{\ln x} \ \ dx$

This is a function of a. Now differentiating w.r.t a on both sides, we get

$I'(a) = \int_0^1 \frac{d}{da} \frac{x^a-1}{\ln x } \ \ dx = \int_0^1 x^a dx = \frac{1}{a+1}$
So we have the equation$I'(a) = \frac{1}{a+1}$ which if we integrate between 0 and a, we get

$\int_0^ 7 I'(a) da= \int_0^7 \frac{1}{a+1} da = \ln 8$

LHS = I(7)-I(0). We can easily see that I(0) = 0. So we get $\boxed{I(7) = \ln 8}$
jagdish singh singh
173 Points
4 years ago
$\hspace{-0.70 cm}Given I = \int^{1}_{0}\frac{x^7-1}{\ln x}dx = \int^{1}_{0}\bigg(\int^{7}_{0}x^{t}dt\bigg)dx\\\\\\ So I =\int^{7}_{0}\bigg(\int^{1}_{0}x^tdx\bigg)dt = \int^{7}_{0}\frac{1}{t+1}dt = \ln(8).$..........................

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