definite integral from 0 to 1 of 5xln(x+1)dx

Answer 1

Use integration by parts. Look for a u which gets simpler in its derivative, and a v which can be integrated ok.

u = ln(x+1)
du = 1/(x+1) dx

dv = 5x dx
v = 5/2 x^2

Int(5x ln(x+1))dx = 5/2 x^2 ln(x+1) - Int(5/2 x^2 * 1/(x+1)) dx

The new integral can be done just using long division:

x^2/(x+1) = x - 1 + 1/(x+1)

5/2 Int(x - 1 + 1/(x+1)) = 5/4 x^2 - 5/2 x + 5/2 ln(x+1)

So, the whole integral is

5/2 x^2 ln(x+1) - (5/4 x^2 - 5/2 x + 5/2 ln(x+1))

f(x) = 5/4(2(x^2 - 1)ln(x+1) - x(x-2))

f(1) = 5/4(2(0)ln2 - 1(-1)) = 5/4
f(0) = 5/4(2(-1)0 - 0(-2)) = 0

So, the answer is 5/4