Answer:
[tex](\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C[/tex]
Step-by-step explanation:
Ok, so we start by setting the integral up. The integral we need to solve is:
[tex]\int x ln(5+x)dx[/tex]
so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:
U=5+x
du=dx
x=U-5
so when substituting the integral will look like this:
[tex]\int (U-5) ln(U)dU[/tex]
now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:
[tex]\int (pq')=pq-\int qp'[/tex]
so we must define p, q, p' and q':
p=ln U
[tex]p'=\frac{1}{U}dU[/tex]
[tex]q=\frac{U^{2}}{2}-5U[/tex]
q'=U-5
and now we plug these into the formula:
[tex]\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU[/tex]
Which simplifies to:
[tex]\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU[/tex]
Which solves to:
[tex]\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C[/tex]
so we can substitute U back, so we get:
[tex]\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C[/tex]
and now we can simplify:
[tex]\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C[/tex]
[tex]\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C[/tex]
[tex]\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C[/tex]
notice how all the constants were combined into one big constant C.
