We have seen how we can find the antiderivative of a function of the form #f(x)=\frac{p(x)}{ax+b}#. We will now have a look at finding the antiderivative of functions of the form #f(x)=\frac{\blue{l \cdot x+m}}{\green{x^2+bx+c}}#.
For these functions, finding the antiderivative is somewhat more complicated and depends on the value of the discriminant of the denominator. If the discriminant is greater than or equal to #0#, we can use fraction decomposition to write the integrand as a sum of simpler integrands. If the discriminant is smaller than #0#, we can use the substitution method by means of completing the square.
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Step-by-step
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Example |
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Determine #\displaystyle \int \frac{\blue{l \cdot x+m}}{\green{x^2+bx+c}} \; \dd x#.
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#\displaystyle \int \frac{\blue{17x+11}}{\green{x^2-2x-3}} \; \dd x# |
Step 1 |
Determine the descriminant of #\green{x^2+bx+c}# using:
\[D=b^2-4c\]
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#D=(-2)^2-4 \cdot -3=16# |
Step 2 |
If #D \lt 0#, go to tab step-by-step 1.
If #D=0#, go to tab step-by-step 2.
If #D \gt 0#, go to tab step-by-step 3.
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#D \gt 0#, go to step-by-step 3 |
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Step-by-step 1
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Example |
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Determine #\displaystyle \int \frac{l \cdot x+m}{x^2+\blue{b}x+c} \; \dd x# with #D \lt 0#.
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#\displaystyle \int \frac{17x+11}{{x^2\blue{-2}x+3}} \; \dd x # |
Step 3 |
Rewrite the integral as:
\[\displaystyle \int \frac{\frac{l}{2}(2x+\blue{b})}{x^2+\blue{b}x+c} \; \dd x + \int \frac{k}{x^2+\blue{b}x+c} \; \dd x\]
Here #k# is equal to #m-\frac{\blue b \cdot l}{2}#, such that #l \cdot x+m=\frac{l}{2}(2x+\blue{b})+k#.
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#\begin{array}{rcl}&&\displaystyle \int \frac{\frac{17}{2}(2x\blue{-2})}{x^2\blue{-2}x+3} \; \dd x \\&+& \displaystyle \int \frac{28}{x^2\blue{-2}x+3} \; \dd x\end{array}# |
Step 4 |
First calculate #\displaystyle \int \frac{\frac{l}{2}(2x+\blue{b})}{x^2+\blue{b}x+c} \; \dd x # by rewriting as
\[\displaystyle \frac{l}{2} \int \frac{1}{x^2+\blue{b}x+c} \; \dd (x^2+\blue{b}x+c)\]
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#\displaystyle \frac{17}{2} \int \frac{1}{x^2\blue{-2}x+3} \; \dd (x^2\blue{-2}x+3)# |
Step 5 |
Using the substitution method, we now calculate that this integral is equal to:
\[\frac{l}{2} \cdot \ln|x^2+\blue b+c|+\green{C_1}\]
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#\frac{17}{2} \cdot \ln|x^2\blue{-2}x+3|+\green{C_1}# |
Step 6 |
Now calculate #\int \frac{k}{x^2+\blue{b}x+c} \; \dd x# by rewriting it as:
\[\int \frac{k}{(x-p)^2+q} \; \dd x\]
Here, #p# and #q \gt 0# are numbers, which we can find by completing the square.
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#\int \frac{28}{(x-1)^2+2} \; \dd x# |
Step 7 |
We now want to rewrite the integral, so that after substitution we can write the integral as #a \cdot \int \frac{1}{x^2+1}#. First, we move #q# in the denominator outside the brackets:
\[\int \frac{k}{q\left(\frac{1}{q}\left(x-p\right)^2+1\right)} \; \dd x\]
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#\int \frac{28}{2 \cdot \left(\frac{1}{2}\left(x-1\right)^2+1\right)} \; \dd x# |
Step 8 |
Now bring the constant in front of the integral. This gives:
\[\frac{k}{q} \int \frac{1}{\frac{1}{q}\left(x-p\right)^2+1} \; \dd x\]
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#\frac{28}{2} \int \frac{1}{\left(\frac{1}{2}\left(x-1\right)^2+1\right)} \; \dd x# |
Step 9 |
Rewrite by placing the factor in front of the square inside the parantheses. This gives:
\[\frac{k}{q} \int \frac{1}{\left(\frac{x-p}{\sqrt{q}}\right)^2+1} \; \dd x\]
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#\frac{28}{2} \int \frac{1}{\left(\frac{x-1}{\sqrt{2}}\right)^2+1} \; \dd x# |
Step 10 |
We can calculate this integral (using substitution of #u=\tfrac{x-p}{\sqrt{q}}#). This gives:
\[\frac{k}{q} \cdot \sqrt{q} \arctan(\frac{x-p}{\sqrt{q}}) +\green{C_2}\]
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#14 \cdot \sqrt{2} \arctan(\frac{x-1}{\sqrt{2}})+\green{C_2}# |
Step 11 |
Now combine the answers from step 6 and step 10 to the final answer of #\displaystyle \int \frac{l \cdot x+m}{x^2+\blue{b}x+c} \; \dd x#. Only one new constant of integration #\green C# is needed.
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#\begin{array}{rcl}&&\frac{17}{2} \cdot \ln|x^2\blue{-2}x+3|\\&+&14 \cdot \sqrt{2} \arctan(\frac{x-1}{\sqrt{2}})+\green{C} \end{array}# |
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Step-by-step 2
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Example |
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Determine #\displaystyle \int \frac{l \cdot x+m}{x^2+bx+c} \; \dd x# with #D = 0#.
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#\displaystyle \int \frac{17x+11}{{x^2-2x+1}} \; \dd x # |
Step 3 |
Rewrite the denominator as #(x+\blue p)^2# for a number #\blue p#. This is possible because #D=0#.
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#(x-\blue{1})^2# |
Step 4 |
Rewrite #\frac{l \cdot x+m}{(x+\blue p)^2}# using fraction decomposition as #\frac{A}{x+\blue p}+\frac{B}{(x+\blue p)^2}#.
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#\frac{17x+11}{(x-\blue{1})^2}=\frac{17}{x-\blue1}+\frac{28}{(x-\blue1)^2}# |
Step 5 |
Decompose the integral to:
\[\displaystyle \int \frac{A}{(x+\blue p)} \; \dd x + \int \frac{B}{(x+\blue p)^2} \; \dd x \]
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#\int \frac{17}{x-\blue1} \; \dd x + \int \frac{28}{(x-\blue1)^2} \; \dd x# |
Step 6 |
Calculate #\int \frac{A}{(x+\blue p)} \; \dd x# as follows:
\[\begin{array}{rcl} \displaystyle \int \frac{A}{(x+\blue p)} \; \dd x &=& \displaystyle A \int \frac{1}{(x+\blue p)} \; \dd x \\ &=& A \ln|x+\blue p|+\green{C_1} \end{array}\]
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#17 \ln|x-\blue1|+\green{C_1} # |
Step 7 |
Calculate #\int \frac{B}{(x+\blue p)^2} \; \dd x # as follows:
\[\begin{array}{rcl} \displaystyle \int \frac{B}{(x+\blue p)^2} \; \dd x &=& \displaystyle \int B \cdot (x+\blue p)^{-2} \; \dd x \\ &=& -\frac{B}{x+\blue p}+\green{C_2} \end{array}\]
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#-\frac{28}{x-\blue1}+\green{C_2}# |
Step 8 |
Now combine the answers from step 6 and step 7 to the final answer of #\displaystyle \int \frac{l \cdot x+m}{x^2+b x+c} \; \dd x#. Only one new constant of integration #\green C# is needed.
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#17 \ln|x-\blue1|-\frac{28}{x-\blue1}+\green{C}# |
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Step-by-step 3
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Example |
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Determine #\displaystyle \int \frac{l \cdot x+m}{x^2+bx+c} \; \dd x# with #D \gt 0#.
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#\displaystyle \int \frac{17x+11}{{x^2-2x-3}} \; \dd x # |
Step 3 |
Rewrite the denominator as #(x+\blue p) \cdot (x+\orange q)# in which #\blue p# and #\orange q# are numbers.
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#(x \blue{-3}) \cdot (x+\orange{1})# |
Step 4 |
Use fraction decomposition to rewrite #\frac{l \cdot x+m}{x^2+bx+c}# as
\[\frac{A}{x+\blue p}+\frac{B}{x+\green q}\]
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\[\frac{17x+11}{x^2-2x-3}= \frac{\frac{31}{2}}{x\blue{-3}}+\frac{\frac{3}{2}}{x+\orange1}\]
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Step 5 |
Now rewrite the integral as:
\[\int \frac{A}{x+\blue p} \; \dd x + \int \frac{B}{x+\orange q} \; \dd x\]
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#\int \frac{\frac{3}{2}}{x\blue{-3}} \; \dd x + \int \frac{\frac{31}{2}}{x+\orange1} \; \dd x# |
Step 6 |
Determine #\int \frac{A}{x+\blue p} \; \dd x#. It holds that:
\[\int \frac{B}{x+\blue p} \; \dd x= A \cdot \ln|x+\blue p| +\green{C_1}\]
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#\int \frac{\frac{31}{2}}{x\blue{-3}} \; \dd x=\frac{31}{2} \ln|x\blue{-3}|+\green{C_1}# |
Step 7 |
Determine #\int \frac{B}{x+\orange q} \; \dd x#.
It holds that:
\[\int \frac{A}{x+\orange q} \; \dd x= B \cdot \ln|x+\orange q| +\green{C_2}\]
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#\int \frac{\frac{3}{2}}{x+\orange1} \; \dd x=\frac{3}{2} \ln|x+\orange{1}|+\green{C_2}# |
Step 8 |
Combine the answers from steps 6 and 7 to the final answer of #\displaystyle \int \frac{l \cdot x+m}{x^2+bx+c} \; \dd x# with #D \gt 0#. Only one new constant of integration #\green C# is needed. |
#\frac{31}{2} \ln|x\blue{-3}|+\frac{3}{2} \ln|x+\orange{1}|+\green{C} # |
Determine #\int {{9-4\cdot x}\over{x^2-8\cdot x+16}} \,\dd x#.
Use #C# to denote the constant of integration.
#\int {{9-4\cdot x}\over{x^2-8\cdot x+16}} \,\dd x=# #-4\cdot \ln \left(\left| x-4\right| \right)+{{7}\over{x-4}}+C#
Because the integrand already has a linear function as numerator of the quotient function, and #a=1# in the denominator, we can start at step 4 of the step-by-step guide.
Step 4 |
We calculate the discriminant of the denominator of the quotient function. \[\begin{array}{rcl}D&=&b^2-4 \cdot c \\ &&\phantom{xxx}\blue{\text{formula discriminant}} \\ &=& (-8)^2-4 \cdot 16 \\ &&\phantom{xxx}\blue{b=-8 \text{ and } c=16 \text{ substituted}} \\ &=& 0 \\ &&\phantom{xxx}\blue{\text{calculated}}\end{array}\] |
Step 5 |
#D =0#, so we move to step-by-step guide 2. |
Step 6 |
We rewrite the denominator of the quotient function. \[x^2-8\cdot x+16=\left(x-4\right)^2\] |
Step 7 |
We rewrite the integral: \[\int {{9-4\cdot x}\over{x^2-8\cdot x+16}} \; \dd x=\int \frac{-4 \cdot (x-4)-7}{\left(x-4\right)^2} \; \dd x\] |
Step 8 |
We split the integral: \[\int \frac{-4 \cdot (x-4)}{\left(x-4\right)^2} \; \dd x + \int \frac{-7}{\left(x-4\right)^2} \; \dd x\] |
Step 9 |
We first calculate #\int \frac{-4 \cdot (x-4)}{\left(x-4\right)^2} \; \dd x# as follows: \[\begin{array}{rcl}\displaystyle \int \frac{-4 \cdot (x-4)}{\left(x-4\right)^2} \; \dd x&=& \displaystyle \int \frac{-4}{x-4} \\ &&\phantom{xxx}\blue{\text{numerator and denominator divided by }x-4} \\&=&-4\cdot \ln \left(\left| x-4\right| \right)+C_1 \\ &&\phantom{xxx}\blue{\text{found the antiderivative}} \end{array}\] |
Step 10 |
\[\int \frac{-7}{\left(x-4\right)^2} \; \dd x = {{7}\over{x-4}}+C_2\] |
Step 11 |
Therefore: \[\int {{9-4\cdot x}\over{x^2-8\cdot x+16}} \,\dd x=-4\cdot \ln \left(\left| x-4\right| \right)+{{7}\over{x-4}}+C\] |