Integral Calculus> Sir can you tell the solution of this cha...

Sir can you tell the solution of this challenging question? The answer given isln 8

Aditya Dev , 8 Years ago

Grade Select Grade

7 Answers

KAPIL MANDAL

Last Activity: 8 Years ago

∫{(x⁷+1) / ln (x) }dx

At first, you put ln(x)=z

This gives, x=e^z and x^-1dx=dz

Then, ∫{(x⁷+1) / ln (x) }dx =∫{(x⁸+x) / z}dz = ∫{(e^8z+e^z) / z}dz.

Now you seperate those.

∫{(e^8z+e^z) / z}dz = ∫{(e^8z) / z}dz + ∫{(e^z) / z}dz. ….(1)

The formula you need to know is,

∫e^cx/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,

∫e^cx/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.

Aditya Dev

Last Activity: 8 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

Last Activity: 8 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

Approved

mycroft holmes

Last Activity: 8 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

Approved

Aditya Dev

Last Activity: 8 Years ago

Yes sir it is minus sign not plus............................................................,............

mycroft holmes

Last Activity: 8 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

Last Activity: 6 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).$$ ..........................