2

ok so i've had a problem trying to simplify the $\ln\left[ \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right]$ and this is supposed to be equal to : $\ln [ \sqrt{a^2+u^2} + u ]$

how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?

Michael Hardy
  • 1
shep
  • 121

3 Answers3

1

I expect that you got this as the result of an (indefinite) integration, and $a$ is a constant. Let $a$ be positive. We are taking the ln of $$\frac{1}{a}\left(\sqrt{u^2+a^2}+u\right).$$ Taking the ln, we get $$\ln\left(\sqrt{a^2+u^2}+u\right)-\ln a.$$ But $\ln a$ is a constant, so can be absorbed into the constant of integration.

In more detail, if $$\ln\left(\sqrt{1+\frac{u^2}{a^2}}+\frac{u}{a}\right)+C$$ is the answer to an indefinite integral problem, where $C$ is an arbitrary constant, then $$\ln\left(\sqrt{a^2+u^2}+u\right)+D$$ is a correct answer to the same problem.

This sort of thing happens a lot, particularly with trigonometric functions. As a simple example, if $\sin^2 x+C$ is "the" answer to an indefinite integration problem, then so is $-\cos^2 x+C'$.

André Nicolas
  • 507,029
  • yes this is true, thnx but the integration requests of me to que rid of - ln qand i don´t know how this is posible, the integration excercise is this: integral of: u^2 * sqrt(a^2+u^2)du and im having trouble with the final part of this integration which is getting rid of - ln (a) – shep Sep 13 '14 at 21:39
  • That gets absorbed into the constant of integration. Or if you are doing a definite integral, the "ugly" version and the nicer version can each be used, since they are both antiderivatives of the function you were integrating. The two functions are not equal, but they differ by a constant. – André Nicolas Sep 13 '14 at 21:55
0

$\ln x$ is injective so, if $\ln x=\ln y\implies x=y$. So, the following should hold true:\begin{align} \sqrt{1+\frac{u^2}{a^2}}+\frac{u}{a}=\sqrt{a^2+u^2}+u\end{align}

However, simply plugging in $u=1, a=2$ gives us

\begin{align*} &\sqrt{1+\frac{1}{4}}+\frac{1}{2}=\sqrt{4+1}+1\\ &\implies \sqrt{\frac{5}{4}}+\frac{1}{2}=\sqrt{5}+1\\ &\implies \sqrt{\frac{5}{4}}=\sqrt{5}-\frac{1}{2}\\ &\implies \frac{\sqrt{5}}{2}=\frac{2\sqrt{5}-1}{2}\\ &\implies \sqrt{5}=2\sqrt{5}-1 \end{align*} which is false.

Hence they are not the same.

Sujaan Kunalan
  • 10,944
0

\begin{align} & \ln\left( \sqrt{1+\frac{u^2}{a^2}} + \frac{u}{a} \right) = \ln\left(\sqrt{\frac{a^2}{a^2}+\frac{u^2}{a^2}} + \frac u a\right) = \ln\left(\sqrt{\frac{a^2+u^2}{a^2}} + \frac u a\right) \\[10pt] = {} & \ln\left(\frac {\sqrt{a^2+u^2}} a + \frac u a\right) = \ln\left(\frac{\sqrt{a^2+u^2}+u} a\right) = \ln\left(\sqrt{a^2+u^2} + u\right) - \ln a \end{align}

If this is viewed as a function of $u$, then $a$ is constant, so it's $$ \ln\left(\sqrt{a^2+u^2} + u\right) + \text{constant} $$

Michael Hardy
  • 1