Thread: Difficult Indefinite Integral

A friend of mine (and now I am) is trying to solve this for a high school kid's AP calculus course. I'm not really worried about the limit part but more how you work it into the correct form to get arctan as an answer. Wolfram took away their free step by step solutions and I'm out of tricks of my own.

$\displaystyle \int_{-\infty }^{\infty }\frac{dx}{(64+x)\sqrt{x}}$

Thanks in advance!

Originally Posted by jjtjp

A friend of mine (and now I am) is trying to solve this for a high school kid's AP calculus course. I'm not really worried about the limit part but more how you work it into the correct form to get arctan as an answer. Wolfram took away their free step by step solutions and I'm out of tricks of my own.
$\displaystyle \int_{-\infty }^{\infty }\frac{dx}{(64+x)\sqrt{x}}$

Have a look at this.

I'm sorry but I don't understand what I am to look at. I know the answer can be had via wolfram, but I'm more interested in the steps taken to arrive at the answer.

Originally Posted by jjtjp

I'm sorry but I don't understand what I am to look at. I know the answer can be had via wolfram, but I'm more interested in the steps taken to arrive at the answer.

$\displaystyle \frac{1}{{\left( {64 + x} \right)\sqrt x }} = \frac{1}{{\left( {64\sqrt x } \right)\left[ {1 + {{\left( {\frac{{\sqrt x }}{8}} \right)}^2}} \right]}}$.

That is clearly arctangent form.

Originally Posted by jjtjp

A friend of mine (and now I am) is trying to solve this for a high school kid's AP calculus course. I'm not really worried about the limit part but more how you work it into the correct form to get arctan as an answer. Wolfram took away their free step by step solutions and I'm out of tricks of my own.

Thanks in advance!

$$\begin{align*}&\int_{-\infty}^{\infty}\frac{1}{64+x}\frac{dx}{\sqrt{x}} \\ \\

&u=\sqrt{x} \\

&du=\frac{dx}{2\sqrt{x}}\Rightarrow 2du=\frac{dx}{\sqrt{x}} \\ \\

&2 \int_{-\infty}^{\infty} \frac{1}{64+u^2}du = \\ \\ \\

&\frac{1}{32} \int_{-\infty}^{\infty} \frac{1}{1+\left( \frac{u}{8} \right)^2} du = \\ \\

&\frac{1}{4}\arctan\left(\frac{u}{8}\right) + C = \\ \\

&\frac{1}{4} \arctan\left(\frac{\sqrt{x}}{8} \right) + C\end{align*}$$

Thank you romsek. I knew it from the start it was some form of substitution. But I was so sure that I had to distribute first after which I was trying to substitute for $\displaystyle x^{\frac{-1}{2}}$ which was obviously getting me stuck. It's so obvious now.

That root x might be a problem with a lower limit of minus infinity. Are we sure that the given limits are correct ?

Originally Posted by BobP

That root x might be a problem with a lower limit of minus infinity. Are we sure that the given limits are correct ?

@BobP, You absolutely right to question the wording in this problem.
If the lower limit were 0 then there is no concern what soever .
If we use |x| instead of x in the question then it all works.

It has been known for at least 150 years that it is not true that every derivative is intergable.
That comes a quite a shock to most mathematics majors as well as a good many mathematics teachers.
The good news is that in the 1950's and 60's, a group of mathematicians found a new definition of the integral that makes the fundamental theorem apply ot all derivatives.

Because my PhD thesis was based on those concepts, I just failed to consider your objection which again is correct!