Science:Math Exam Resources/Courses/MATH101/April 2005/Question 03 (a)

MATH101 April 2005

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[hide] Question 03 (a)
Full-Solution Problem. Justify your answer and show all your work. Simplification of your answer is not required. Evaluate the following integral $\displaystyle \int {\frac {4x+4}{x(x+1)^{2}}}\,dx.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Can you rewrite the integrand in a way that will allow you to compute the integral?

[show] Hint 2
Start by simplifying the integrand, then use partial fractions.

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[show] Solution 1
To begin with, let's simplify the integrand: ${\begin{aligned}\int {\frac {4x+4}{x(x+1)^{2}}}\,dx&=4\int {\frac {x+1}{x(x+1)^{2}}}\,dx\\&=4\int {\frac {1}{x(x+1)}}\,dx\end{aligned}}$ Next we use partial fractions to solve ${\frac {1}{x(x+1)}}={\frac {A}{x}}+{\frac {B}{x+1}}$ . Multiplying both sides by $x(x+1)$ we obtain $1=A(x+1)+Bx=(A+B)x+A$ , and when we compare coefficients we find that $A+B=0$ , $A=1$ and hence $B=-1$ . Therefore, ${\begin{aligned}4\int {\frac {1}{x(x+1)}}\,dx&=4\int \left({\frac {A}{x}}+{\frac {B}{x+1}}\right)\,dx\\&=4\int \left({\frac {1}{x}}-{\frac {1}{x+1}}\right)\,dx\\&=4(\ln \|x\|-\ln \|x+1\|+C)\end{aligned}}$

[show] Solution 2
Here is a longer, alternative solution that does not simplify the integrand first. Instead, we will use the method of partial fractions immediately. First, we will rewrite the integrand: ${\frac {4x+4}{x(x+1)^{2}}}={\frac {4x}{x(x+1)^{2}}}+{\frac {4}{x(x+1)^{2}}}.$ We can now cancel a factor of $x$ , leaving: ${\frac {4}{(x+1)^{2}}}+{\frac {4}{x(x+1)^{2}}}.$ We have made a little progress. We have shown that ${\frac {4x+4}{x(x+1)^{2}}}={\frac {4}{(x+1)^{2}}}+{\frac {4}{x(x+1)^{2}}}.$ We want to compute $\int {\frac {4x+4}{x(x+1)^{2}}}dx.$ By the above computation, this integral is equal to $\int {\frac {4}{(x+1)^{2}}}+{\frac {4}{x(x+1)^{2}}}dx,$ which is equal to $\int {\frac {4}{(x+1)^{2}}}dx+\int {\frac {4}{x(x+1)^{2}}}dx,$ by basic properties of the integral. Now we can see why breaking the integrand up into pieces was helpful: the integral to the left of the "+" can be evaluated by making the substitution $u=x+1,du=dx$ . Thus, $\int {\frac {4}{(x+1)^{2}}}dx=\int {\frac {4}{u^{2}}}du=-4u^{-1}+C=-4(x+1)^{-1}+C.$ Now we will have to use partial fractions to compute the next integral. This means we need to find constants A, B, and C to make the following equality true: ${\frac {4}{x(x+1)^{2}}}={\frac {A}{x}}+{\frac {B}{x+1}}+{\frac {C}{(x+1)^{2}}}.$ Multiply both sides by $x(x+1)^{2}$ to cancel the denominators, yielding $4=A(x+1)^{2}+Bx(x+1)+Cx.$ Now we expand the right hand side. $4=A(x^{2}+2x+1)+Bx^{2}+Bx+Cx$ $4=Ax^{2}+2Ax+A+Bx^{2}+Bx+Cx$ $4=(A+B)x^{2}+(2A+B+C)x+A$ Now since we have an equality of polynomials, we know that their coefficients of $x^{2}$ and $x$ must be equal, as well as the constant coefficient. The polynomial on the left hand side has coefficient of $x^{2}$ and $x$ equal to zero, and constant coefficient equal to 4. Thus we get three equations: For the coefficient of $x^{2}$ : $0=A+B$ For the coefficient of $x$ : $0=2A+B+C$ For the constant coefficient: $4=A$ Since A=4, we get from the $x^{2}$ coefficient equation that $0=4+B$ Thus, B=-4. Now we can solve for C in the $x$ coefficient equation, $0=2*4+(-4)+C$ Thus, C=-4. Putting all of this together, we can conclude that ${\frac {4}{x(x+1)^{2}}}={\frac {4}{x}}+{\frac {-4}{x+1}}+{\frac {-4}{(x+1)^{2}}}.$ So: $\int {\frac {4}{x(x+1)^{2}}}dx=\int {\frac {4}{x}}dx+\int {\frac {-4}{x+1}}dx+\int {\frac {-4}{(x+1)^{2}}}dx.$ We have succeeded in splitting up a difficult integral into three simpler ones! $\int {\frac {4}{x}}dx=4\ln \|x\|+c_{1},$ $\int {\frac {-4}{x+1}}dx=-4\ln \|x+1\|+c_{2},$ $\int {\frac {-4}{(x+1)^{2}}}dx=4(x+1)^{-1}+c_{3},$ where $c_{1},c_{2},c_{3}$ are constants. Thus $\int {\frac {4}{x(x+1)^{2}}}dx=4\ln \|x\|-4\ln \|x+1\|+4(x+1)^{-1}+C$ Finally, we are able to conclude that ${\begin{aligned}\int {\frac {4x+4}{x(x+1)^{2}}}dx=-4(x&+1)^{-1}+4\ln \|x\|\\&-4\ln \|x+1\|+4(x+1)^{-1}+C\end{aligned}}$ where the term $-4(x+1)^{-1}$ comes from the integral we did at the very beginning. Notice two terms cancel out to get as our final answer $\int {\frac {4x+4}{x(x+1)^{2}}}{\textrm {d}}x=4\ln \|x\|-4\ln \|x+1\|+C.$