# In = integral of x^n/(x+1), x=0 to x=1. What is the relation between the terms In+1 and In?

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Student Comments

giorgiana1976 | Student

We'll calculate first, I0.

I0 = Int x^0dx/(x+1)

I0 = Int dx/(x+1)

I0 = ln(x+1)

We'll apply Leibniz-Newton to evaluate the value of definite integral I0.

I0 = F(1) - F(0)

I0 = ln(1+1) - ln(0+1)

I0 = ln2 - ln1, but ln 1 = 0

I0 = ln2

We'll determine I1:

I1 = Int x^1dx/(x+1)

I1 = Int xdx/(x+1)

I1 = Int dx - Int dx/(x+1)

But Int dx/(x+1) = I0 = ln 2

I1 = x - I0

I1 = 1 - 0 - I0

I1 = 1 - ln 2

We notice that In+1 = Int x^ndx - In

In+1 + In = Int x^ndx

In+1 + In = x^(n+1)/(n+1), for x = 0 to x = 1

**The relation between the consecutive terms is In+1 + In = 1/(n+1).**