Solving an Integral with Integration by Parts

[fredgarvin22]

hello

could someone give me a pointer here.

this integral
∫ln(x + c)dx

my guess is, by integration by parts
(ab)' = a'b + ab'
∫ba = ab - ∫b'a

so here
a = ln(c + x) b = c + x
a' = 1/(c + x) b' = 1

ab = (c + x)*ln(c + x)
and
∫b'a = ∫ ((c + x)/(hc + x)) dx
= ∫dx = x
so ab - ∫b'a = (c + x)*ln(c + x) - x


would this be correct?

 

[quasar987]

I think you messed up a little bit in your re-writting the problem, but the answer (c + x)*ln(c + x) - x is correct. When it comes to integrals, you can always verify your answer by differentiating your answer. If it gives the integrand, you've got the right answer. If not, there's a mistake.

 

[tacman]

Yes, your approach is correct. To solve this integral using integration by parts, you can choose a = ln(c + x) and b = c + x, which gives you a' = 1/(c + x) and b' = 1. Then using the formula for integration by parts, you can evaluate the integral as (c + x)*ln(c + x) - ∫(c + x)(1/(c + x))dx = (c + x)*ln(c + x) - x + C. So, your final answer is (c + x)*ln(c + x) - x + C. I hope this helps!