1
$\begingroup$

In my math book it says that it is practical to know the following primitive, a is any real number. $$\int \frac 1 {\sqrt{x^2 + a}} \, dx = \ln |x + \sqrt{x^2 + a}|+ C$$

There is no more explanation about this, so i was wondering if someone could explain this to me.

What im trying to do is following.

$$\int \frac 1 {\sqrt{x^2 + a}} \, dx = \int (x^2 + a)^{-1/2} \, dx = \frac{(x^2 + a)^{1/2}} x + C$$

In the last step i use the power formula for the derivative and the inner derivative which give me $2x$. I cant see were im doing it wrong.

  • 0
    You have to substitute $u = x^2 + a$ because power formula works only if there's no constant. To check if your answer is correct, just take the derivative. You should find that the derivative of $(x^2 + a)^{1/2} x^{-1}$ is actually $\frac{a}{x^2 \sqrt{x^2 - a}}$. – 2017-01-30
  • 1
    A formula tells you that $$ \int x^n \, dx = \frac{x^{n+1}}{n+1}+C, $$ but that does not mean that $$ \int(\text{some function of }x)^n \,dx = \frac{(\text{that same function})^{n+1}}{n+1} + C. $$ One problem is that if you let $u = \text{that function of }x,$ then $du$ is not the same as $dx.$ Another problem is that when you divided by $x$ you're doing something with $x$ that can only be validly done with constants, i.e. quantities not depending on $x.\qquad$ – 2017-01-30
  • 0
    So i use a variable change: $ [ u = x^2 +a => x = \sqrt (u-a)]\,[ \frac{dx}{du} = \frac {1}{2\sqrt(u-a)} => dx = \frac {du}{2\sqrt(u-a)}] \qquad \int \frac{1}{\sqrt{x^2 + a}} dx =\int \frac{1}{2u\sqrt{u-a}} du = ?$ From here i wonder what to do. Shall i use the partial integration formula? After i rewrite $\int \frac{1}{2u\sqrt{u-a}} du = \int (2u)^{-1}*(u-a)^{-1/2} du$ – 2017-01-30

1 Answers 1

1

Emerald, What you are trying to do, is not going to work because it is incorrect. You cannot "account for the chainrule" by simply dividing by $x$ because it is a variable, not some coefficient. One way to integrate given integral is through u-sub $x=a\tan t$ and thus $dx=a\sec^2t\,dt$. This result in integrating one single term $a\sec t$. Now there is a standard anti derivative for that one, but if you want to find explanations for it, it certainly can be found somewhere on this site. Can you give it a try from here?

  • 0
    Thank you for your answer, that cleared it somewhat for me. I did the misstake of dividing with x get rid of the other x in the numerator. I just wanted to know a some more about the acctual standard anti derivate. So i could use it even tho the anti derivate doesnt look excatly the same. For example $\int \frac{1}{\sqrt{t^2+2t+3}} dt $ or something similar. – 2017-01-30
  • 0
    In this last case just complete the square under the square root. – 2017-01-30
  • 1
    Yeah i figured that out: $\int \frac{-1}{(t+1)^2+2} dt $ I also seen that that given anti derivate equals $ -ln (t + 1 + \sqrt{(t+1)^2+2}) + C$ The problem is that i just fail to see the answer from using the standard anti derivate $ \int \frac 1 {\sqrt{x^2 + a}} \, dx = \ln |x + \sqrt{x^2 + a}|+ C $ According to my mathbook i should easily see the answer if i use this standard anti derivate. – 2017-01-30