# Integral of $1/[(1+x^2)\sqrt{1+x^2}]$

I try to get back on track with the integration. I would like to solve

$$\int_0^1 \frac{dx}{(1+x^2)\sqrt{1+x^2}}.$$

There are my way to try to solve it (that I don't find the right solution) and an other way proposed in the book that I don't understand :

Answer : $\frac{1}{\sqrt{2}}$.

1. My way : $$\int_0^1 \frac{dx}{(1+x^2)\sqrt{1+x^2}} = \int_0^1 (1+x^2)^{\frac{-3}{2}}dx$$ With the substitution : $t = 1+x^2$ $$\int_0^1 (1+x^2)^{-\frac{3}{2}}dt = \int_1^2 2tt^{-3/2}dt = 2\int_1^2 t^{-1/2}dt = 4\sqrt{t}|^2_1 = 4\sqrt{2}-3$$ which is wrong.
I don't know if I did something that I wasn't allowed.

2. Book's way.
With the substitution : $x = \tan(t)$ $$\int_0^{\pi/4} \frac{1+\tan^2(t)}{(1+\tan^2(t))\sqrt{1+\tan^2(t)}}dt = \int_0^{\pi/4} \cos(t) dt = \sin(t)|^{\pi/4}_0 = \frac{1}{\sqrt{2}}.$$ How did they find that was equal to $\cos(t)$ ? How did they find that would be a good idea to substitute with $\tan$ ?

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$$t=1+x^2 \Rightarrow dt = 2x \,dx$$ Therefore $$\int_0^1 (1+x^2)^{-\frac{3}{2}}dt \neq \int_1^2 2tt^{-3/2}dt$$ – etothepitimesi Aug 17 '14 at 12:54
Oh, so I totally inverted ? To keep this example, I should use $$t=x+\frac{x^3}{3}$$ and then $$dt=1+x^2$$? (I know this is stupid, just for the example) – XogoX Aug 17 '14 at 12:57
you're missing a $dx$ in your last expression, but I think you got the idea. $$t=x+\frac{x^3}{3} \Rightarrow dt = \left( 1+x^2 \right)dx \Rightarrow dx = \frac{dt}{1+x^2}$$ – etothepitimesi Aug 17 '14 at 12:59
Ok perfect ! Now it rings a bell – XogoX Aug 17 '14 at 13:02
This seems related to this question. I derive the primitive, $\dfrac{x}{\sqrt{1+x^2}}$, in my answer, which for some reason was downvoted, so beware ;-). – robjohn Aug 17 '14 at 20:01

First note that:

$$I=\int_{0}^{1}\frac{\mathrm{d}x}{(1+x^2)\sqrt{1+x^2}}=\int_{0}^{1}\frac{\mathrm{d}x}{(1+x^2)^{3/2}}.$$

If you then had the presence of mind to substitute $u=\frac{x}{\sqrt{1+x^2}}$, you would notice that

$$\mathrm{d}u=\frac{\mathrm{d}x}{(1+x^2)^{3/2}},$$

and thus,

$$I=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}u=\frac{1}{\sqrt{2}}.$$

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These are some kind of standard trigonometric substitutions: $$\sqrt{x^2-a^2}\text{ or }x^2-a^2\tag{x=a\sin\theta or x=a\cos\theta}$$ $$\sqrt{x^2+a^2}\text{ or }x^2+a^2\tag{x=a\tan\theta or x=a\cot\theta}$$

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@XogoX sorry i just posted incomplete answer accidently, see now – ADG Aug 17 '14 at 12:59

1.If $t=1+x^2$ then $dt=2xdx$not $dx=2tdt$ 2.A very useful rule of thumb is to substitute $x=tan t$ when you see $\sqrt{1+x^2}$ or $1/(1+x^2)$ to use $1+tan^2t=sec^2t$.Similar tricks include using x=sin/cos for $\sqrt{1-x^2}$ like terms and x=sec/csc for $\sqrt{x^2-1}$ like terms.

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Is there a list of rules of thumb that I can found for the integration ? – XogoX Aug 17 '14 at 13:01
– Vignesh Manoharan Aug 17 '14 at 13:09

By using the trigonometric substitution $x=\tan(t)$, we can see how the integrand simplifies to $\cos(t)$ $\frac{(1+\tan^{2}(t))}{(1+\tan^{2}(t))\sqrt{1+\tan^{2}(t)}} =\frac{1}{\sqrt{1+\tan^{2}(t)}}=\cos(t)$ I hope this helps you understand.

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(+1) since this is essentially how I evaluated the primitive in this answer. – robjohn Aug 17 '14 at 20:19

However, this integral doesn't need more attention; you could also use the substitution to solve the indefinite associated integral:

$$1+x^2=t^2x^2$$

And soyou'll find the integral as $$\int(-1/t^2)dt$$

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