007B Sample Midterm 1, Problem 4 Detailed Solution

Evaluate the following integrals.

(a)   $\int x^{2}e^{x}~dx$

(b)   $\int {\frac {5x-7}{x^{2}-3x+2}}~dx$

---

Background Information:
1. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$
2. Through partial fraction decomposition, we can write the fraction
${\frac {1}{(x+1)(x+2)}}={\frac {A}{x+1}}+{\frac {B}{x+2}}$
for some constants $A,B.$

Solution:

(a)

Step 1:
We proceed using integration by parts.
Let  $u=x^{2}$  and  $dv=e^{x}dx.$
Then,  $du=2xdx$  and  $v=e^{x}.$
Therefore, we have
$\int x^{2}e^{x}~dx=x^{2}e^{x}-\int 2xe^{x}~dx.$

Step 2:
Now, we need to use integration by parts again.
Let  $u=2x$  and  $dv=e^{x}dx.$
Then,  $du=2dx$  and  $v=e^{x}.$
Building on the previous step, we have
${\begin{array}{rcl}\displaystyle {\int x^{2}e^{x}~dx}&=&\displaystyle {x^{2}e^{x}-{\bigg (}2xe^{x}-\int 2e^{x}~dx{\bigg )}}\\&&\\&=&\displaystyle {x^{2}e^{x}-2xe^{x}+2e^{x}+C.}\end{array}}$

(b)

Step 1:
We need to use partial fraction decomposition for this integral.
Since  $x^{2}-3x+2=(x-1)(x-2),$  we let
${\frac {5x-7}{(x-1)(x-2)}}={\frac {A}{x-1}}+{\frac {B}{x-2}}.$
Multiplying both sides of the last equation by  $(x-1)(x-2),$
we get
$5x-7=A(x-2)+B(x-1).$
If we let  $x=2,$  the last equation becomes  $3=B.$
If we let  $x=1,$  then we get  $-2=(-1)A.$  Thus,  $A=2.$
So, in summation, we have
${\frac {5x-7}{x^{2}-3x+2}}={\frac {2}{x-1}}+{\frac {3}{x-2}}.$

Step 2:
Now, we have
$\int {\frac {5x-7}{x^{2}-3x+2}}~dx=\int {\frac {2}{x-1}}~dx+\int {\frac {3}{x-2}}~dx.$
Now, we use  $u$-substitution to evaluate these integrals.
For the first integral, we substitute  $u=x-1.$
For the second integral, the substitution is  $t=x-2.$
Then, we integrate to get
${\begin{array}{rcl}\displaystyle {\int {\frac {5x-7}{x^{2}-3x+2}}~dx}&=&\displaystyle {\int {\frac {2}{u}}~du+\int {\frac {3}{t}}~dt}\\&&\\&=&\displaystyle {2\ln |u|+3\ln |t|+C}\\&&\\&=&\displaystyle {2\ln |x-1|+3\ln |x-2|+C.}\end{array}}$

Final Answer:
(a)     $x^{2}e^{x}-2xe^{x}+2e^{x}+C$
(b)     $2\ln |x-1|+3\ln |x-2|+C$

Return to Sample Exam