# int_0^1(x^7-1)/lnxdx = ?

May 4, 2018

$I = \ln \left(8\right)$

#### Explanation:

We want to solve

$I = {\int}_{0}^{1} \frac{{x}^{7} - 1}{\ln} \left(x\right) \mathrm{dx}$

The indefinite integral involves the exponential- and logarithmic integral, so we will take a different approach

This will be solved by differentiation under the integral sign

Introduce a parameter color(red)(a, and consider

$I \left(a\right) = {\int}_{0}^{1} \frac{{x}^{a} - 1}{\ln} \left(x\right) \mathrm{dx}$

Notice, we seek color(blue)(I(7), and moreover that color(blue)(I(0)=0

Differentiate both sides with respect to color(red)(a

$I ' \left(a\right) = {\int}_{0}^{1} \frac{d}{\mathrm{da}} \left(\frac{{x}^{a} - 1}{\ln} \left(x\right)\right) \mathrm{dx}$

$\textcolor{w h i t e}{I ' \left(a\right)} = {\int}_{0}^{1} {x}^{a} \mathrm{dx}$

$\textcolor{w h i t e}{I ' \left(a\right)} = {\left[{x}^{a + 1} / \left(a + 1\right)\right]}_{0}^{1}$

$\textcolor{w h i t e}{I ' \left(a\right)} = \frac{1}{a + 1}$

Integrate both sides with respect to color(red)(a

$I \left(a\right) = \int \frac{1}{a + 1} \mathrm{da} = \ln \left(a + 1\right) + C$

But color(blue)(I(0)=0, so the constant can evaluated

$0 = \ln \left(0 + 1\right) + C \implies C = 0$

Thus

$I \left(a\right) = \ln \left(a + 1\right)$

For color(red)(a=7

$I \left(7\right) = \ln \left(7 + 1\right) = \ln \left(8\right)$

Or equivalent

$I = {\int}_{0}^{1} \frac{{x}^{7} - 1}{\ln} \left(x\right) \mathrm{dx} = \ln \left(8\right)$