Science:Math Exam Resources/Courses/MATH101/April 2009/Question 04 (b)

MATH101 April 2009

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[hide] Question 04 (b)
Calculate the integral: $\int {\sqrt {(1+e^{x})}}dx$

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.

[show] Hint 1
Try a substitution first.

[show] Hint 2
Try the substitution $u={\sqrt {1+e^{x}}}$

[show] Hint 3
Then run through the partial fractions argument. It might help simplify things to notice that $2u^{2}=2u^{2}-2+2$

[show] Hint 4
Don't forget to plug back in for x and your +C

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.
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[show] Solution
Following the hint, we start off by setting $u={\sqrt {1+e^{x}}}$ Then we have that $\displaystyle u^{2}-1=e^{x}$ and more importantly that $du={\frac {e^{x}}{2{\sqrt {1+e^{x}}}}}\,dx={\frac {u^{2}-1}{2u}}\,dx$ Thus, substituting these values in yield $\int {\sqrt {1+e^{x}}}\,dx=\int u\left({\frac {2u}{u^{2}-1}}\,du\right)=\int {\frac {2u^{2}}{u^{2}-1}}\,du$ Via the third hint, we can simplify this last integral to $\int {\frac {2u^{2}}{u^{2}-1}}\,du=\int {\frac {2u^{2}-2+2}{u^{2}-1}}\,du=\int 2\,du+\int {\frac {2}{u^{2}-1}}\,du$ (one could find this expression by explicitly doing the long division though on an exam you probably want to try to save some time). The first integral is $\int 2\,du=2u=2{\sqrt {1+e^{x}}}$ (we'll add the constant in later) and the second integral we evaluate by partial fractions. ${\frac {2}{u^{2}-1}}={\frac {2}{(u+1)(u-1)}}={\frac {A}{u+1}}+{\frac {B}{u-1}}={\frac {A(u-1)+B(u+1)}{(u+1)(u-1)}}$ This gives the expression $\displaystyle 2=A(u-1)+B(u+1)$ Plugging in $u=1$ yields $\displaystyle 2=A(1-1)+B(1+1)=2B$ and so $B=1$ . Plugging in $u=-1$ yields $\displaystyle 2=A((-1)-1)+B((-1)+1)=-2A$ and so $A=-1$ . Hence $\int {\frac {2}{u^{2}-1}}\,du=\int -{\frac {1}{u+1}}\,du+\int {\frac {1}{u-1}}\,du=-\ln \|u+1\|+\ln \|u-1\|$ (we'll add the constant in later). Hence, applying everything, we have ${\begin{aligned}\int {\sqrt {1+e^{x}}}\,dx&=\int {\frac {2u^{2}}{u^{2}-1}}\,du\\&=\int 2\,du+\int {\frac {2}{u^{2}-1}}\,du\\&=2u-\ln \|u+1\|+\ln \|u-1\|+C\\&=2{\sqrt {1+e^{x}}}-\ln \|{\sqrt {1+e^{x}}}+1\|+\ln \|{\sqrt {1+e^{x}}}-1\|+C\end{aligned}}$ completing the question.