Question

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

Answer 1

`I=ln(8)`

Explanation:

We want to solve

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

`I(a)=int_0^1(x^a-1)/ln(x)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'(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)`