$\int_{0}^{1} \frac{x^{7} - 1}{ln x} d x$ $int_0^1(x^7-1)/lnxdx$ = ?

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Answer 1 · 2018-05-04T08:00:26

$I = ln (8)$ $I=ln(8)$

We want to solve

$I = \int_{0}^{1} \frac{x^{7} - 1}{ln (x)} d x$ $I=int_0^1(x^7-1)/ln(x)dx$

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 $a$ $color(red)(a$ , and consider

$I (a) = \int_{0}^{1} \frac{x^{a} - 1}{ln (x)} d x$ $I(a)=int_0^1(x^a-1)/ln(x)dx$

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

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

$I'(a)=int_0^1d/(da)((x^a-1)/ln(x))dx$

$color(white)(I'(a))=int_0^1x^adx$

$color(white)(I'(a))=[x^(a+1)/(a+1)]_0^1$

$color(white)(I'(a))=1/(a+1)$

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

$I(a)=int1/(a+1)da=ln(a+1)+C$

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

$0=ln(0+1)+C=>C=0$

Thus

$I(a)=ln(a+1)$

For $color(red)(a=7$

$I(7)=ln(7+1)=ln(8)$

Or equivalent

$I=int_0^1(x^7-1)/ln(x)dx=ln(8)$

int_0^1(x^7-1)/lnxdx∫10x7−1lnxdxint_0^1(x^7-1)/lnxdx = ?