Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 64

From the basic list of power series, we have:
ln(x) =sum_(n=0)^oo (-1)^(n) (x-1)^(n+1)/(n+1)
= (x-1)-(x-1)^2/2+(x-1)^3/3 -(x-1)^4/4 +...
We replace "x " with "x+1 " to setup:
ln(1+x) =sum_(n=0)^oo (-1)^n ((x+1)-1)^(n+1)/(n+1)
=sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)
=x-x^2/2+x^3/3 -x^4/4+...
Note: ((x+1)-1) = (x+1-1) = x
Then,
x ln(1+x) =sum_(n=0)^oo (-1)^n x^(n+1)/(n+1) *x
=sum_(n=0)^oo (-1)^n x^(n+2)/(n+1)

Note: x^(n+1) * x = x^(n+1+1) =x^(n+2)
Applying the summation formula, we get:
x ln(1+x)= x*[x-x^2/2+x^3/3 -x^4/4+...]
or
= x^2 -x^3/2+x^4/3-x^5/4 +...
Then the integral becomes:
int_0^(1/4) xln(x+1) = int_0^(1/4) [x^2 -x^3/2+x^4/3-x^5/4 +...]dx
To determine the indefinite integral, we integrate each term using the Power Rule for integration: int x^n dx= x^(n+1)/(n+1) .
int_0^(1/4) [x^2 -x^3/2+x^4/3-x^5/4 +...]dx
= [x^3/3 -x^4/(2*4)+x^5/(3*5)-x^6/(4*6) +...]_0^(1/4)
= [x^3/3 -x^4/7+x^5/15-x^6/24 +...]_0^(1/4)
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
F(1/4) or F(0.25) =0.25^3/3 -0.25^4/7+0.25^5/15-0.25^6/24 +...
=1/192-1/1792+1/15360 -1/98304+...
F(0)=0^3/3 -0^4/7+0^5/15-0^6/24 +...
= 0-0+0-0+...
All the terms are 0 then F(0) =0 .
We may stop at 4th term (1/98304~~0.00001017) since we only need an error less than 0.0001 .
F(1/4)-F(0) = [1/192-1/1792+1/15360 -1/98304]-[0]
= 0.00470522926
Thus, the approximated integral value:
int_0^(1/4) xln(x+1) dx ~~0.0047