Integration: Integration techniques
Integration by parts
We have already seen how we can use the substitution method, based on the chain rule, for integration. Now we will look at a method to integrate based on the product rule.
Iintegration by parts
We can apply integration by parts for two functions #\blue{g(x)}# and #\orange{h(x)}#:
\[
f(x) = \blue {g(x)} \cdot \green{h'(x)}\]
gives
\[\begin{array}{rcl}\displaystyle \int f(x) \; \dd x&=&\displaystyle\int \blue {g(x)} \cdot \green{h'(x)} \dd x \\ &=&\blue{g(x)} \cdot \orange{h(x)} - \displaystyle \int \purple{g'(x)} \cdot \orange{h(x)} \; \dd x
\end{array}\]
Example
#\begin{array}{rcl}&&\displaystyle \int x \e^{2x} \; \dd x \\ &=&\displaystyle\int \blue {x} \cdot \green{\e^{2x}} \dd x \\ &=&\blue{x} \cdot \orange{\dfrac{1}{2}\e^{2x}} - \displaystyle \int \purple{1} \cdot \orange{\frac{1}{2}\e^{2x}} \; \dd x\end{array}#
Step-by-step guide integration by parts
Step-by-step | Example | |
Determine #\displaystyle \int f(x) \; \dd x# using integration by parts. |
Determine #\displaystyle \int x \e^{2x} \dd x# | |
Step 1 |
Determine #\blue{g(x)}# and #\green{h'(x)}# such that #f(x)=\blue {g(x)} \cdot \green{h'(x)}#. |
#\blue{g(x)}=x# #\green{h'(x)}=\e^{2x}# |
Step 2 |
Determine #\purple{g'(x)}# and #\orange{h(x)}#. |
#\purple{g'(x)}=1# #\orange{h(x)}=\frac{1}{2} \e^{2x}# |
Step 3 |
Calculate #\int f(x) \; \dd x# with: \[\begin{array}{rcl}\displaystyle \int f(x) \; \dd x&=&\displaystyle\int \blue {g(x)} \cdot \green{h'(x)} \dd x \\ &=&\blue{g(x)} \cdot \orange{h(x)} - \displaystyle \int \purple{g'(x)} \cdot \orange{h(x)} \; \dd x \end{array}\] |
\[\blue{x} \cdot \orange{\frac{1}{2} \e^{2x}} - \displaystyle \int \purple{1} \cdot \orange{\frac{1}{2} \e^{2x}} \; \dd x \] |
Step 4 |
Determine #\int \purple{g'(x)} \cdot \orange{h(x)} \; \dd x#. |
#\displaystyle \int \purple{1} \cdot \orange{\frac{1}{2} \e^{2x}} \; \dd x= \frac{1}{4}\e^{2x}+C_1# |
Step 5 |
Determine the final answer by substituting step 4 in step 3. |
# \frac{1}{2} x \cdot \e^{2x}-\frac{1}{4} \e^{2x}+C# |
Step 1 | We are looking for #g(x)# and #h'(x)#, such that #x\cdot \e^ {- 4\cdot x }=g(x) \cdot h'(x)#. In this case we choose #g(x)=x# and #h'(x)=\e^ {- 4\cdot x }#. |
Step 2 | We now calculate #g'(x)# and #h(x)#. #g'(x)=1# #h(x)=-{{\e^ {- 4\cdot x }}\over{4}}# |
Step 3 | According to the calculation rule for integration by parts: \[\int g(x) \cdot h'(x) \; \dd x=g(x) \cdot h(x) - \int g'(x) \cdot h(x) \; \dd x\] This gives: \[\int x\cdot \e^ {- 4\cdot x } \,\dd x=x \cdot -{{\e^ {- 4\cdot x }}\over{4}} - \int 1 \cdot -{{\e^ {- 4\cdot x }}\over{4}} \; \dd x\] |
Step 4 | We now calculate: #\int 1 \cdot -{{\e^ {- 4\cdot x }}\over{4}} \; \dd x# \[\int 1 \cdot -{{\e^ {- 4\cdot x }}\over{4}} \; \dd x={{\e^ {- 4\cdot x }}\over{16}}+C_1\] |
Step 5 | We now substitute step 4 in step 3 in. That gives \[\int x\cdot \e^ {- 4\cdot x } \,\dd x=-{{x\cdot \e^ {- 4\cdot x }}\over{4}}-{{\e^ {- 4\cdot x }}\over{16}} + C\] |
Or visit omptest.org if jou are taking an OMPT exam.