Compute the definite integral:
integral_0^1 x log(x+1) dx
For the integrand x log(x+1), substitute u = x+1 and du = dx.
This gives a new lower bound u = 1+0 = 1 and upper bound u = 1+1 = 2:
= integral_1^2 (u-1) log(u) du
Expanding the integrand (u-1) log(u) gives u log(u)-log(u):
= integral_1^2 (u log(u)-log(u)) du
Integrate the sum term by term and factor out constants:
= integral_1^2 u log(u) du- integral_1^2 log(u) du
For the integrand u log(u), integrate by parts, integral f dg = f g- integral g df, where
f = log(u), dg = u du,
df = 1/u du, g = u^2/2:
= 1/2 u^2 log(u)|_1^2-1/2 integral_1^2 u du- integral_1^2 log(u) du
Evaluate the antiderivative at the limits and subtract.
1/2 u^2 log(u)|_1^2 = 1/2 2^2 log(2)-1/2 1^2 log(1) = log(4):
= log(4)-1/2 integral_1^2 u du- integral_1^2 log(u) du
Apply the fundamental theorem of calculus.
The antiderivative of u is u^2/2:
= log(4)+(-u^2/4)|_1^2- integral_1^2 log(u) du
Evaluate the antiderivative at the limits and subtract.
(-u^2/4)|_1^2 = (-2^2/4)-(-1^2/4) = -3/4:
= -3/4+log(4)- integral_1^2 log(u) du
For the integrand log(u), integrate by parts, integral f dg = f g- integral g df, where
f = log(u), dg = du,
df = 1/u du, g = u:
= -3/4+log(4)+(-u log(u))|_1^2+ integral_1^2 1 du
Evaluate the antiderivative at the limits and subtract.
(-u log(u))|_1^2 = (-(2 log(2)))-(-log(1)) = -log(4):
= -3/4+ integral_1^2 1 du
Apply the fundamental theorem of calculus.
The antiderivative of 1 is u:
= -3/4+u|_1^2
Evaluate the antiderivative at the limits and subtract.
u|_1^2 = 2-1 = 1:
Answer: | = 1/4
