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

KAPIL MANDAL
132 Points
							∫{(x7+1) / ln (x) }dxAt first, you put ln(x)=zThis gives, x=ez and x-1dx=dzThen, ∫{(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) = 0Now you have only the first part.Now simple algebra will lead you to the result. Any problem you face, write here.

3 years ago
28 Points
							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!) ?

3 years ago
KAPIL MANDAL
132 Points
							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

3 years ago
mycroft holmes
272 Points
							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

3 years ago
28 Points
							Yes sir it is minus sign not plus............................................................,............

3 years ago
mycroft holmes
272 Points
							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 $\boxed{I(7) = \ln 8}$

3 years ago
jagdish singh singh
173 Points
							$\hspace{-0.70 cm}Given I = \int^{1}_{0}\frac{x^7-1}{\ln x}dx = \int^{1}_{0}\bigg(\int^{7}_{0}x^{t}dt\bigg)dx\\\\\\ So I =\int^{7}_{0}\bigg(\int^{1}_{0}x^tdx\bigg)dt = \int^{7}_{0}\frac{1}{t+1}dt = \ln(8).$.......................... …...................................................................... ….......................................................................

one year ago
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• 51 Video Lectures
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• Test paper with Video Solution
• Mind Map
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• NCERT Solutions
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• Previous Year Exam Questions