I'm reviewing for an intro calculus exam, and the following problem appears on a past final exam:
If: $$\int^3_1 \frac{1}{x^4\sqrt{1+x}}\, dx = k$$
What is:
$$\int^3_1 \frac{1}{x^5\sqrt{1+x}}\, dx $$ (I'm assuming the answer will be in terms of $k$)
It seems that most basic integration techniques(substitution, integration by parts, trig sub, etc.) will not allow the solution of the integral, and I'm not sure how else to approach this problem at my level. I've run this by both my lecturers, and they cannot find a solution in a reasonable amount of time either. I'm curious because it seems the there would be a simple solution or rule I'm ignorant of (considering this is on an intro calc exam), but I'm stumped. Where am I going wrong? Thanks!
Answer
Let $I_4$ be the given integral (i.e. k) and let $I_5$ be the integral you want. I am just expanding on Zach Stone's comment, so credit due to him, one can intgrate by parts and write
$$I_4 = \int^3_1 \frac{1}{x^4}\frac{1}{\sqrt{1+x}}\, dx $$ which will give
$$I_4 = \left. \frac{2\sqrt{1+x}}{x^4}\right|_1^3 + \int^3_1 \frac{8\sqrt{1+x}}{x^5}dx$$
The trick is now to multiply and divide the second term within the integral by $\sqrt{1+x}$. The equation then easily simplifies to
$$I_4 = \frac{4}{81} - 2\sqrt{2} +8I_4 +8I_5$$ which gives the required relation between $I_4$ and $I_5$. I verified that it agrees with Brian Tung's calculations from Wolfram.
So, in short, yes, this was possible using simple integration by parts. Once again, credit to Zach Stone
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