Can you evaluate #int \ (xln(x+1))/(x+1) \ dx #?

Question

Scarlett Allison · Accepted Answer

I tried this:

Have a look:

Isabella Ackerman · Answer

# int \ (xln(x+1))/(x+1) \ dx = (x+1)ln(x+1) - x - 1/2 ln^2(x+1) + C #

Yes I can! We want to find: # I = int \ (xln(x+1))/(x+1) \ dx # An obvious logical substitution would be: Let #w=x+1 = > (dw)/dx = 1#, and #x=w=1# Substituting into the integral gives us: # I = int \ ((w-1)lnw)/w \ dw # # \ \ = int \ (w lnw - ln w)/w \ dw # # \ \ = int \ lnw - ln w/w \ dw # # \ \ = int \ lnw \ dw - int \ ln w/w \ dw # The first integral is a well known result and one that probably should be memorised. It can be derived using Integration by parts: Let # { (u,=lnw, => , (du)/(dw)=1/w), ((dv)/(dw),=1, =>, v=w ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx # gives us: # int \ (lnw)(1) \ dw = (lnw)(w) - int \ (w)(1/1) \ dw # # :. int \ lnw \ dw = wlnw - int \ dw # # :. " " = wlnw - w # And for the second integral we can perform another substitution: Let #z=lnw => (dz)/(dw) = 1/w# Substituting into the second integral gives us: # int \ ln w/w \ dw = int \ z \ dz # # " " = 1/2z^2 # And restoring the substitution for #w# we get # int \ ln w/w \ dw = 1/2 (lnw)^2 # Combining these two results then gives us: # I = {wlnw - w} - {1/2 (lnw)^2} + C_1 # And restoring the substitution for #w# gives us: # I = (x+1)ln(x+1) - (x+1) - 1/2 ln^2(x+1) + C_1 # # \ \ = (x+1)ln(x+1) - x-1 - 1/2 ln^2(x+1) + C_1 # # \ \ = (x+1)ln(x+1) - x - 1/2 ln^2(x+1) + C #

Samantha Adkison · Answer

undefined To evaluate the integral $\int \frac{x\ln(x+1)}{x+1} \, dx$, we can use integration by parts.

Let $u = \ln(x+1)$ and $dv = \frac{x}{x+1} \, dx$. Then, $du = \frac{1}{x+1} \, dx$ and $v = \int \frac{x}{x+1} \, dx$.

To find $v$, we can perform polynomial long division or notice that $x = (x+1) - 1$, so we can rewrite the integral as $\int \left(1 - \frac{1}{x+1}\right) \, dx$.

Integrating term by term, we get $v = x - \ln(x+1)$.

Now, we use the formula for integration by parts: $\int u \, dv = uv - \int v \, du$.

Substituting our values in, we have:

$$
\begin{aligned}
\int \frac{x\ln(x+1)}{x+1} \, dx &= x\ln(x+1) - \int (x - \ln(x+1)) \cdot \frac{1}{x+1} \, dx \
&= x\ln(x+1) - \int \frac{x}{x+1} \, dx + \int \frac{\ln(x+1)}{x+1} \, dx \
&= x\ln(x+1) - (x - \ln(x+1)) + C \
&= x\ln(x+1) - x + \ln(x+1) + C,
\end{aligned}
$$

where $C$ is the constant of integration.