Integration: Integration techniques
Finding the antiderivatives of quotient functions 1
We will now check how we can use long division to find the antiderivative of a quotient function of the form #f(x)=\frac{p(x)}{ax+b}# where #p(x)# is a polynomial and #a# and #b# are numbers.
For a polynomial #\blue{q(x)}# and numbers #a#, #b# and #m#, the following is true:
\[\int \blue{q(x)}+\frac{m}{ax+b} \; \dd x= \orange{Q(x)} +\frac{m}{a}\ln|ax+b|+\green C\]
Here #\orange{Q(x)}# is the antiderivative of #\blue{q(x)}# and #\green C# is the constant of integration.
Example
#\begin{array}{rcl}&&\displaystyle \int \blue{x^2+\frac{1}{2}x-2 }+\frac{3}{2x+2}\; \dd x \\&=&\orange{\frac{1}{3}x^3+\frac{1}{4}x^2-2x}+\frac{3}{2}\ln|2x+2|+\green C\end{array}#
To find the antiderivative of a function in the form #\frac{p(x)}{ax+b}#, we rewrite it first using long division. We then use the rule explained above.
Step-by-step |
Example | |
Determine #\displaystyle \int \frac{\blue{p(x)}}{\green{ax+b}} \; \dd x#, where #p(x)# is a polynomial. |
#\displaystyle \int \frac{\blue{2x^3+3x^2-3x-1}}{\green{2x+2}} \; \dd x# | |
Step 1 |
Rewrite #\frac{\blue{p(x)}}{\green{ax+b}}# as #q(x)+\frac{m}{\green{ax+b}}# using long division, where #q(x)# is a polynomial and #m# is a number. |
#x^2+\frac{1}{2}x-2+\frac{3}{\green{2x+2}}# |
Step 2 |
Rewrite: \[\begin{array}{rcl} && \displaystyle \int \frac{\blue{p(x)}}{\green{ax+b}} \; \dd x\\&=&\displaystyle \int q(x) \; \dd x+\displaystyle \int \frac{m}{\green{ax+b}} \; \dd x\end{array}\] |
#\begin{array}{rcl}&&\displaystyle \int \frac{\blue{2x^3+3x^2-3x-1}}{\green{2x+2}} \; \dd x \\ &=& \displaystyle \int x^2+\frac{1}{2}x-2 \; \dd x+\displaystyle \int \frac{3}{\green{2x+2}}\; \dd x \end{array}# |
Step 3 |
Determine #\int q(x) \; \dd x#. |
#\frac{1}{3}x^3+\frac{1}{4}x^2-2x# |
Step 4 |
Determine #\int \frac{m}{\green{ax+b}} \; \dd x#. |
#\frac{3}{2}\ln|\green{ax+b}|# |
Step 5 |
Determine #\displaystyle \int \frac{\blue{p(x)}}{\green{ax+b}} \; \dd x# by adding step 3 and step 4. |
#\frac{1}{3}x^3+\frac{1}{4}x^2-2x+\frac{3}{2}\ln|\green{ax+b}|+C# |
Step 1 | We rewrite the integrand in a polynomial and a fraction of the form #\frac{r}{ax+b}#, where #r#, #a# and #b# are numbers (using long division): \[{{8\cdot x-2}\over{8\cdot x+3}}=1+\frac{-5}{8\cdot x+3}\] |
Step 2 | We can now split the integral: \[\int {{8\cdot x-2}\over{8\cdot x+3}} \; \dd x=\int 1 \; \dd x+\int \frac{-5}{8\cdot x+3} \; \dd x\] |
Step 3 | We first calculate #\int 1 \; \dd x# using the rule of calculation for finding the antiderivative of a power function (#1=1 \cdot x^0#). \[\int 1 \; \dd x=x+C_1\] |
Step 4 | We now calculate #\int \frac{-5}{8\cdot x+3} \; \dd x#. This gives: \[\begin{array}{rcl}\displaystyle \int \frac{-5}{8\cdot x+3} \; \dd x&=& \displaystyle -5 \cdot \int \frac{1}{8\cdot x+3} \; \dd x \\ &&\phantom{xxx}\blue{\text{moved constant in front of integral}} \\&=& \displaystyle -5 \int \frac{1}{8} \cdot \frac{1}{8\cdot x+3} \; \dd(8\cdot x+3) \\ &&\phantom{xxx}\blue{\text{rewritten so we can substitute}} \\ &=& \displaystyle \frac{-5}{8} \int \frac{1}{8\cdot x+3} \; \dd(8\cdot x+3) \\ &&\phantom{xxx}\blue{\text{moved constant outside of integral}} \\&=& \displaystyle \frac{-5}{8} \int \frac{1}{u} \; \dd u \\ &&\phantom{xxx}\blue{\text{substituted }8\cdot x+3=u} \\ &=& \displaystyle \frac{-5}{8} \ln(|u|) +C_2 \\ &&\phantom{xxx}\blue{\text{found antiderivative}} \\ &=& \displaystyle -{{5\cdot \ln \left(\left| 8\cdot x+3\right| \right)}\over{8}} +C_2 \\ &&\phantom{xxx}\blue{\text{substituted }u=8\cdot x+3 \text{ and if possible simplified constant}} \end{array}\] |
Step 5 | We now combine the answers to step 3 and 4 to find the final answer. \[\int {{8\cdot x-2}\over{8\cdot x+3}} \; \dd x=x-{{5\cdot \ln \left(\left| 8\cdot x+3\right| \right)}\over{8}}+C\] |
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