0
$\begingroup$

So I tried evaluating the following indefinite integral and I did two solutions: $$\int \sqrt{1+\frac 3x}dx$$ 1. My first solution was to use a substitution $u=1+\frac3x$ and my approach went as follows: $$\int \sqrt{1+\frac 3x}dx\,\begin{vmatrix}u=1+\frac3x \\du=-\frac {3}{x^2}dx\end{vmatrix}\,=\int \frac{-\sqrt{u}x^2}{3}du=-3\int \frac{\sqrt{u}}{(u-1)^2}du\,\begin{vmatrix}v=\sqrt{u} \\dv=\frac {1}{2\sqrt{u}}du\end{vmatrix}\\ =-6\int \frac{v^2}{(v^2-1)^2}dv$$ This integral comes out to be $$\frac{3v}{(v^2-1)}-\frac{3}{2}(\ln|v-1|-\ln|v+1|)$$ Substituting back in terms of u's and then x's gives the final answer of $$x\sqrt{1+\frac{3}{x}}-\frac{3}{2}\left(\ln\left\lvert\sqrt{1+\frac{3}{x}}-1\right\rvert-\ln\left\lvert\sqrt{1+\frac{3}{x}}+1\right\rvert\right)+c$$ 2. My second solution was to use the trigonometric substitution $x=3\cot^2(\theta)$ and my solution was: $$\int \sqrt{1+\frac{3}{x}}dx\,\begin{vmatrix}x=3\cot^2(\theta) \\dx=-6\cot(\theta)\csc^2(\theta)d\theta\end{vmatrix}\,=-6\int \sqrt{1+\tan^2(\theta)}\cot(\theta)\csc^2(\theta)d\theta\\ =-6\int \sec{\theta}\cot(\theta)\csc^2(\theta)d\theta=-6\int \csc^3(\theta)d\theta$$ Using the reduction formula $$\int \csc^n(\theta)d\theta=-\left(\frac{1}{n-1}\right)\csc^{n-2}(\theta)\cot(\theta)+\left(\frac{n-2}{n-1}\right)\int \csc^{n-2}(\theta)d\theta$$ , this integral comes out to be $$3\csc(\theta)\cot(\theta)+3\ln|\csc(\theta)+\cot(\theta)|$$ After setting up a right triangle and substitution back in for x's I end up with $$\sqrt{x^2+3x}+3 \ln|\sqrt{x+3}+\sqrt{x}|+c$$ When I graphed both answers, I realized that my first solution yielded a graph that was defined for along the entire domain of the original function , while my second solution yielded a graph that was only defined along the positive domain of the original function. So my question is, why does this occur, whether or not I am missing something, and how do I determine the approach that would guarantee an antiderivative defined along the entire domain of the original function? Any feedback would be appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

When you make the substitution $x=3\cot^2\theta,$ you guarantee that $x$ will be nonnegative.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.