Evaluate the integral of x*ln(1+x) dx by first making a substitution and then using integration by parts.
I let u=1+x and du = dx but then ln(u) du can't integrate to anything?
To evaluate the integral ∫x*ln(1+x)dx, we can make the substitution u = 1+x. Let's calculate the differentials:
du = dx.
Now we need to express the entire integral in terms of u. We can rewrite x in terms of u as x = u - 1. Substituting x = u - 1 and dx = du into the integral, we have:
∫[(u - 1)*ln(u)]du.
Now, to evaluate this integral using integration by parts, let's choose u = ln(u) and dv = (u - 1)du.
Taking the derivatives and integrals, we get:
du = du, and v = ∫(u - 1)du = (u^2/2 - u) + C,
where C is the constant of integration.
Now applying the integration by parts formula:
∫u*dv = uv - ∫v*du,
we have:
∫[(u - 1)*ln(u)]du = ln(u) * [(u^2/2 - u)] - ∫[(u^2/2 - u)*du].
Simplifying, we get:
∫[(u - 1)*ln(u)]du = (u^2/2 - u)ln(u) - ∫(u^2/2 - u)du.
Expanding and simplifying the expression, we have:
∫[(u - 1)*ln(u)]du = (u^2/2 - u)ln(u) - (u^3/6 - u^2/2) + C.
Finally, substituting u = 1 + x back into the expression, we obtain:
∫x*ln(1+x) dx = [((1+x)^2/2 - (1+x))ln(1+x) - ((1+x)^3/6 - (1+x)^2/2)] + C.
This expression is the evaluation of the given integral using substitution and integration by parts.
you need to get rid of that pesky ln(1+x). So,
u = ln(1+x)
du = 1/(1+x) dx
dv = x dx
v = 1/2 x^2
∫u dv = uv - ∫v du
= 1/2 x^2 ln(1+x) - 1/2 ∫x^2/(1+x) dx
which is not so hard, eh?