How do you integrate $(x+2)\ln(x-3)$? I got $$\left(\frac {x^2}{2} +2x\right)\ln(x-3)-\left(\frac {x^2}{4}-\frac {7x}{2} -\frac {21}{2}\right)\ln(2x-6)$$ as my answer... Not sure If I got it right. Please correct me, thank you!
 A: Hint. You may use an integration by parts
$$
\begin{align}
\int(x+2)\ln(x-3)\:dx&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+2)^2}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\frac{(x+7)(x-3)+25}{x-3}\:dx
\\\\&=\frac12(x+2)^2\ln (x-3)-\frac12\int\left(x+7+\frac{25}{x-3}\right)\:dx.
\end{align}
$$ Then conclude easily.
A: $\displaystyle \int (x+2)\ln(x-3)dx=\int x\ln(x-3)dx+\int 2\ln(x-3)dx$
Note that we make use of the formulas $\displaystyle\int \ln u du=u\ln u-u+C$ and $\displaystyle\int u\ln udu=\frac{1}{4}u^2(2\ln u-1)+C$, which can be shown via integration by parts.
For the first integral, let $u=x-3$ and $du=dx$ 
This yields $\displaystyle\int (u+3)\ln udu=\int (u\ln u+3\ln u )du=\frac{1}{4}u^2(2\ln u-1)+3u\ln u-u+C_1=\frac{1}{4}(x-3)^2(2\ln (x-3)-1)+3(x-3)\ln (x-3)-(x-3)+C_1$
For the second integral, let $u=x-3$ and $du=dx$
This yields $\displaystyle2\int \ln u du=2u\ln u-2u+C_2=2(x-3)\ln (x-3)-2(x-3)+C_2.$
Therefore $$\displaystyle \int (x+2)\ln(x-3)dx=\frac{1}{4}(x-3)^2(2\ln (x-3)-1)+3(x-3)\ln (x-3)-(x-3)+2(x-3)\ln (x-3)-2(x-3)+C$$
You can further simplify the antiderivative, but this would be the general strategy for integration.
A: It should be
$$-x^2/4-7x/2+(1/2 x^2 +2x) \ln(x-3)-21/2\ln(3-x)$$
A: $$\int(x+2)\ln(x-3)\space\text{d}x=\int(x\ln(x-3)+2\ln(x-3))\space\text{d}x=$$
$$\int x\ln(x-3)\space\text{d}x+2\int\ln(x-3)\space\text{d}x=$$

Substitute $u=x-3$ and $\text{d}u=\text{d}x$

$$\int(u+3)\ln(u)\space\text{d}u+2\int\ln(u)\space\text{d}u=$$
$$\int(u\ln(u)+3\ln(u))\space\text{d}u+2\int\ln(u)\space\text{d}u=$$
$$\int u\ln(u)\space\text{d}u+3\int\ln(u)\space\text{d}u+2\int\ln(u)\space\text{d}u$$
Now, to find $\int u\ln(u)\space\text{d}u$ use integration by parts, with:
$$\int f\space\text{d}g=fg-\int g\space\text{d}f$$
Where $f=\ln(u),\text{d}g=u\space\text{d}u,\text{d}f=\frac{1}{u}\space\text{d}u,g=\frac{u^2}{2}$
Now, to find $\int\ln(u)\space\text{d}u$ use integration by parts, with:
$$\int f\space\text{d}g=fg-\int g\space\text{d}f$$
Where $f=\ln(u),\text{d}g=\text{d}u,\text{d}f=\frac{1}{u}\space\text{d}u,g=u$
A: With $u=\ln(x-3)$, $du = \frac{1}{x-3} dx$, $dv = (x-3) dx$, $v = \left(\frac{x^2}{2}+2x\right)$,
$$ \int (x+2) \ln(x-3) dx = \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{x-3} \left(\frac{x^2}{2}+2x\right) dx \\
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{w} \left(\frac{(w+3)^2}{2}+2(w+3)\right) dw \\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \frac{1}{w} \left(\frac{w^2+6w+9}{2}+2w+6)\right) dw \\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \int \left( \frac{1}{2}w + 5 +\frac{21}{2w} \right)dw \\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \frac{w^2}{4} -5 w - \frac{21}{2} \ln w + C\\ 
= \left(\frac{x^2}{2}+2x\right) \ln(x-3) - \frac{(x-3)^2}{4} -5 (x-3) - \frac{21}{2} \ln (x-3) + C\\ 
= \left(\frac{x^2}{2}+2x - \frac{21}{2}\right) \ln(x-3) - \frac{(x-3)^2}{4}  -5x + C$$
EDIT:  Here's how you can check if your answer is right.  
A: The integration should proceed as follows:
$$\color{violet}{I=\int (x+2)\ln(x-3)dx}$$
Assume $x-3=z$ and then we have that $dx=dz$
The integrand hence becomes $$\color{indigo}{I=\int (z+5)\ln z dz}$$
$$\color{blue}{=\int z\ln z dz+\int 5\ln z dz}$$
Integrating the first one by parts and then the second one as usual,
$$\begin{align} & \color{green}{I=\ln z\int z dz - \int\{\frac{d}{dz}(\ln z)\int z dz\}dz+5\cdot\left[\ln z\int  dz - \int\{\frac{d}{dz}(\ln z)\int dz\}dz\right]}
\\ & \color{brown}{=\frac{z^2}{2}\ln z - \int \frac{1}{z}\cdot \frac{z^2}{2} dz + 5\left[z\ln z -z\right]}
\\ & \color{orange}{=\frac{z^2}{2}\ln z - \frac{z^2}{4} + 5\left[z\ln z -z\right] + c}
\\ & \color{red}{=\frac{(x-3)^2}{2}\ln (x-3) - \frac{(x-3)^2}{4} + 5\left[(x-3)\ln (x-3) -(x-3)\right] + c}\end{align}$$
A: For convenience we first shift $x-3=t$ to integrate $(t+5)\ln(t)$.
Then by parts,
$$\begin{align}\int(t+5)\ln(t)dt&=\left(\frac{t^2}2+5t\right)\ln(t)-\int \left(\frac{t}2+5\right)dt\\
&=\left(\frac{t^2}2+5t\right)\ln(t)-\left(\frac{t^2}4+5t\right)\\
&=\frac{x^2+4x-21}2\ln(x-3)-\frac{x^2+14x-51}4.\end{align}$$
A: Here's a slightly sneaky way to do the integral, based on knowing (partly from experience) what kind of function, when differentiated, leaves a polynomial times a logarithm.
The answer must be of the form
$$(Ax+B)(x-3)\ln(x-3)-\int(Ax+B)dx$$
where the derivative of $(Ax+B)(x-3)$ is $x+2$.  But the derivative of $(Ax+B)(x-3)$ is $A(x-3)+(Ax+B)=2Ax+(B-3A)$, so $2A=1$ and $B-3A=2$, which imply $A={1\over2}$ and $B=2+3A={7\over2}$. Thus, knowing that $\int(Ax+B)dx={1\over2}Ax^2+Bx+C$, we have
$$\int(x+2)\ln(x-3)dx={1\over2}(x+7)(x-3)\ln(x-3)-{1\over4}x^2-{7\over2}x+C$$
This approach works in general:
$$\int p(x)\ln(x-a)dx=P(x)(x-a)\ln(x-a)-\int P(x)dx$$
where the equality $P'(x)(x-a)+P(x)=p(x)$ determines the coefficients of $P$.
A: Set $\ln(x-3)=t$, so $x=e^t+3$ and $dx=e^t\,dt$. Therefore the integral is
$$
\int(e^t+5)te^t\,dt=\int t(e^{2t}+5e^t)\,dt=
t\left(\frac{1}{2}e^{2t}+5e^t\right)-
\int\left(\frac{1}{2}e^{2t}+5e^t\right)dt
$$
Can you go on?
In the back substitution, note that $e^{2t}=(x-3)^2$.
