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A friend asked me what was the solution to the problem (which was on his test)$$\int\frac4xdx$$ I proceeded to tell him that you can take out the 4 in the numerator, and then just take the integral of $1/x$, and multiply by 4, ending up with an answer of $$4\ln(x) + C$$ He then asked me if you could multiply the fraction by 4 on the top and the bottom, resulting in $$\int\frac{16}{4x}dx$$ Then this equation has a solution of $$4\ln(4x) + C = 4\ln(x) + 8\ln 2 + C$$ He then asked me why his solution is different than mine. I proceeded to tell him that the $+$ $C$ term in both equations is a constant, and to just visualize that the second equation has a different $C$ term than the first one. He then asked me how to explain this to his Calculus AB teacher, to which I had no answer. Any suggestions?

MJD
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Vishwa Iyer
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    Your question is “how can I explain this to a Calculus AB teacher”, correct? – MJD Feb 10 '14 at 16:59
  • This does not answer your question, but the most general function $F(x)$ such that $F'(x)=\frac{4}{x}$ everywhere that $\frac{4}{x}$ is defined is given by $F(x)=4\ln(x)+C$ when $x\gt 0$, and $F(x)=4\ln(-x)+D$ when $x\lt 0$. – André Nicolas Feb 10 '14 at 16:59
  • @MJD Yes. That's my question. – Vishwa Iyer Feb 10 '14 at 17:01
  • You might be interested in this question, which is related. The OP calculated $\int x\sin^2 x;dx$ by two methods, obtaining $\frac14x^2-\frac x4\sin(2x) + \frac14\sin^2 x + C$ by one method and $\frac14x^2-\frac x4\sin(2x) - \frac18\cos(2x) + C$ by the other; he then asked which is correct. But both are correct, because $\frac14\sin^2 x$ and $- \frac18\cos(2x)$ differ by a constant. – MJD Feb 10 '14 at 17:11
  • Here's another, in which $\int -\frac1{2x};dx$ is calculated by two methods and found to be both $-\frac12\ln\lvert x\rvert + C$ and $-\frac12\ln\lvert-2x\rvert + C$. – MJD Feb 10 '14 at 17:13

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Both approach are correct. In fact, in the first one you found: $$4\ln(x)+C_1$$ and in the second one you found: $$4\ln(x)+C_2$$ These two constants play in such a way that we wouldn't face to any contradictions. If you are allowed to speak about initial conditions for an ODE, then we would able to make clearer what is going on here.

Mikasa
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Let $F(x)$ be an antiderivative of $f(x)$. Then $G(x)$ is an antiderivative of $f(x)$ if and only if there exists a constant $C$ such that $F(x)-G(x)=C$. In this problem, we have \begin{align*} F(x) &= 4\cdot\ln|x| \\ G(x) &= 4\cdot\ln|x|+8\cdot\ln2 \\ f(x) &=\frac{4}{x} \end{align*} Clearly $F^\prime(x)=f(x)$ and $F(x)-G(x)=8\cdot\ln2$, so $G(x)$ is also an antiderivative of $f(x)$.

Brian Fitzpatrick
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