I have to solve this

$$\int \frac{dx}{(x+1)\sqrt{x^2+2x}}$$

First make the substitution $u=x+1$ and get

$$\int \frac{ds}{u\sqrt{u^2-1}}$$

Next I substitute $s=\sqrt{u^2-1}$ and I get $ds=u\,du/\sqrt{u^2-1}$.

I have integral of $1/s^2+1\dots$ I replace for $s$ and I get the integral as

$$\arctan\sqrt{u^2-1} \, .$$

The answer on my textbook is $-\arcsin[1/(x+1)]$

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    $\begingroup$ U n r e a d a b l e. $\endgroup$ – Git Gud Jan 20 '13 at 18:27
  • $\begingroup$ try putting in latex $\endgroup$ – Santosh Linkha Jan 20 '13 at 18:33
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    $\begingroup$ @GitGud U n h e l p f u l $\endgroup$ – Fly by Night Jan 20 '13 at 18:40
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    $\begingroup$ @GitGud I see... $\endgroup$ – Fly by Night Jan 20 '13 at 18:42
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    $\begingroup$ @GitGud: for the new user, your comment is not at all helpful. experimentX at least suggested a link that leads to helpful information. $\endgroup$ – robjohn Jan 20 '13 at 19:26

You wanted $\;s=\sqrt{u^2-1}$, so $\;s^2=u^2-1$, $\;2s\,ds=2u\,du$, and $\;s\,ds=u\,du$.

Substituting gives us: $$ \begin{align}\int\frac{du}{u\sqrt{u^2-1}} &= \int\frac{u\,du}{u^2 \sqrt{u^2-1}} \\&= \int\frac{s\,ds}{(s^2+1)s}\\&= \int\frac{ds}{s^2+1} \\ \\&= \arctan s + C. \end{align} $$

Now don't forget to "back substitute": $s = \sqrt{u^2-1} = \sqrt{(x+1)^2 - 1} = \sqrt{x^2 + 2x}$

So our answer is: $$\quad\arctan(\sqrt{x^2 + 2x}) + C$$

Note that with trig substitution, and trigonometric solutions, there can be any number of solutions (constants may vary), as is evident by the number of trigonometric identities! E.g.:

$$\arctan\sqrt{x^2+2x}-\left(\arcsin\left(\frac1{x+1}\right)-\frac\pi2\right) = 0$$

$$-\arcsin\frac1{x+1}=\arctan\sqrt{x^2+2x}-\text{ constant} = \arctan \sqrt{x^2 + 2x} + C$$

To check your integration answers, you can always take the the derivative of your solutions, and if you can derive the original integrand, you are "good to go." Try differentiating each solution (your's and the text's) and both will give you the equivalent of your original integrand.

  • $\begingroup$ @Yves - It you expand those factors $(x+1)$ etc, and simplify, I'm willing to bet that d/dx of each function will be equivalent, through identities. It just amounts to algebraic manipulation and trig identities. $\endgroup$ – Namaste Jan 20 '13 at 23:48
  • $\begingroup$ yes it's exactly the same.... however if I plot the two functions in Wolfram|Alpha (not their derivatives) they seem quite different... $\endgroup$ – Yves Jan 21 '13 at 0:42
  • $\begingroup$ Yves Because they differ by constants, that can "throw off" the graphs. Integrals yield "families of functions/solutions"... $\endgroup$ – Namaste Jan 21 '13 at 0:46
  • $\begingroup$ Here they should differ only by one real constant (assuming we are in R), but it seems more complicated, maybe because the two functions are not defined for the same values ? wolframalpha.com/input/… $\endgroup$ – Yves Jan 21 '13 at 1:05

If $s=\sqrt{u^2-1}$ then $s^2=u^2-1$ so $2s\,ds=2u\,du$, and $s\,ds=u\,du$.

$$ \int\frac{du}{u\sqrt{u^2-1}} = \int\frac{u\,du}{u^2 \sqrt{u^2-1}} = \int\frac{s\,ds}{(s^2+1)s} = \int\frac{ds}{s^2+1} = \arctan s + \text{constant}. $$

Now draw a right triangle in which the "opposite" side has length $\sqrt{u^2-1}$ and the "adjacent" side has length $1$. By the Pythagorean theorem, the hypotenuse has length $u$, and you see the trigonometric identity $$ \arctan\sqrt{u^2-1} = \arcsin\frac{\sqrt{u^2-1}}{u} $$


After your work, there are several possibilities. For example, we can let $v=\frac{1}{u}$. Then $u=\frac{1}{v}$ and $du=-\frac{1}{v^2}\,dv$. Do the rest of the substituting, noting that $\frac{1}{\sqrt{u^2-1}}=\frac{v}{\sqrt{1-v^2}}$. Quickly we arrive at $$\int -\frac{dv}{\sqrt{1-v^2}},$$ which is familiar.

A more standard substitution is $u=\sec\theta$. Then $du=\sec\theta\tan\theta$, and $\sqrt{u^2-1}=\tan\theta$. Everything cancels!


Let $u=1/t$, and get

$$\int \frac{du}{u\sqrt{u^2-1}}=-\int \frac{dt}{\sqrt{1-t^2}}=-\arcsin t +C.$$


Michael Hardy's on the right track here, but I think the answer is incomplete. First of all, it's sloppy to leave that u in there. Your answer in terms of $x$ is $\tan^{-1}\sqrt{x^2+2x}$. As Michael Hardy suggested, create the right triangle, marking one angle. Let the length of the side opposite that angle be $\sqrt{x^2+2x}$ and the adjacent side have a length of $1$. That gives the hypoteneuse a length of $x+1$.

Now, you will see that the angle opposite the angle you marked has a sine of $\frac1{x+1}$. These 2 angles add up to $\frac\pi2$ radians, so you have



Including the constant of integration in your answer, $\tan^{-1}\sqrt{x^2+2x}+C$, we see that the 2 answers are equivalent. Both forms are correct.


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