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

MATH101 April 2005

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Question 03 (b)
Full-Solution Problem. Justify your answer and show all your work. Simplification of your answer is not required. Evaluate the following integral $\displaystyle \int (x+1)\ln(x)\,dx.$

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.

Hint 1
Try making a substitution, and writing the old variable in terms of the new one.

Hint 2
After your first substitution you will find an integral that you can solve using integration by parts.

Hint 3
As an alternative solution, write this integral as the sum of two simpler integrals and use integration by parts for both of them.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We want to evaluate this integral: $\int (x+1)\ln(x)\,dx$ We start by making the following substitution: $y=\ln(x).$ Then $dy\,={\frac {1}{x}}\,dx$ and thus $dx=x\,dy$ So, after making the substitution, the integral becomes $\int (x+1)yx\,dy.$ But because $y=\ln(x),$ it follows that $x=e^{y}.$ So the integral reduces to $\int e^{y}(e^{y}+1)y\,dy$ $=\int (e^{2y}+e^{y})y\,dy$ Now, we apply integration by parts with $u=y$ $du=dy$ $dv=e^{2y}+e^{y}$ $v={\frac {1}{2}}e^{2y}+e^{y}$ So, $\int u\,dv=uv-\int v\,du$ $=y({\frac {1}{2}}e^{2y}+e^{y})-\int {\frac {1}{2}}e^{2y}+e^{y}\,dy$ $=y({\frac {1}{2}}e^{2y}+e^{y})-{\frac {1}{4}}e^{2y}-e^{y}+C.$ $=y({\frac {1}{2}}(e^{y})^{2}+e^{y})-{\frac {1}{4}}(e^{y})^{2}-e^{y}+C$ Finally, we substitute y = ln(x) back in: $={\frac {1}{2}}x^{2}\ln(x)+x\ln x-{\frac {1}{4}}x^{2}-x+C$ So the integral is given by $\int (x+1)\ln(x)dx={\frac {1}{2}}x^{2}\ln(x)+x\ln x-{\frac {1}{4}}x^{2}-x+C$ This is not required in a complete solution, but it is recommended to check your answer by differentiating: ${\begin{aligned}&{\frac {d}{dx}}\left({\frac {1}{2}}x^{2}\ln(x)+x\ln(x)-{\frac {1}{4}}x^{2}-x+C\right)\\&\quad =x\ln(x)+{\frac {1}{2}}x+\ln(x)+1-{\frac {1}{2}}x-1\\&\quad =x\ln(x)+\ln(x)\\&\quad =(x+1)\ln(x).\end{aligned}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. As an alternative solution we rewrite the integral as $\int (x+1)\ln x\,dx=\int x\ln x\,dx+\int \ln x\,dx$ and solve either of the simpler integrals separately. In the first integral, use integration by parts with $u=\ln x$, $du=(1/x)\,dx$ and $dv=x\,dx$, $v=x^{2}/2$ so that ${\begin{aligned}\int x\ln x\,dx&=\int u\,dv\\&=uv-\int v\,du\\&={\frac {1}{2}}x^{2}\ln x-\int \left({\frac {1}{2}}x^{2}\right){\frac {1}{x}}\,dx\\&={\frac {1}{2}}x^{2}\ln x-{\frac {1}{4}}x^{2}+C_{1}\end{aligned}}$ We use a similar integration by parts in the second integral, namely $u=\ln x$, $du=(1/x)\,dx$ and $dv=dx$, $v=x$, which leads to ${\begin{aligned}\int \ln x\,dx&=x\ln x-\int x\left({\frac {1}{x}}\right)\,dx=x\ln x-x+C_{2}\end{aligned}}$ Putting the pieces together and setting $\displaystyle C=C_{1}+C_{2}$, we obtain the final answer $\int x\ln x\,dx={\frac {1}{2}}x^{2}\ln x-{\frac {1}{4}}x^{2}+x\ln x-x+C$

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