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

Question

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

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.
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 $\tan$ ?

$$t=1+x^2 \Rightarrow dt = 2x ,dx$$ 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
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

score 6 · Accepted Answer · answered Aug 17 '14 at 13:03

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,

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

score 1 · Answer 2 · answered Aug 17 '14 at 12:54

1

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

answered Aug 17 '14 at 12:54

RE60K

17,716

1

@XogoX sorry i just posted incomplete answer accidently, see now – RE60K Aug 17 '14 at 12:59

score 1 · Answer 3 · answered Aug 17 '14 at 12:56

1

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.

answered Aug 17 '14 at 12:56

VigneshM

408

Is there a list of rules of thumb that I can found for the integration ? – XogoX Aug 17 '14 at 13:01
1

try Integration tips – VigneshM Aug 17 '14 at 13:09

k170 · Answer 4 · 2014-08-17T13:35:37.820

1

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.

edited Aug 17 '14 at 13:35

answered Aug 17 '14 at 13:19

k170

9,045

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

score 1 · Answer 5 · answered Aug 17 '14 at 18:58

1

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 $$\int(-1/t^2)dt$$

answered Aug 17 '14 at 18:58

Mikasa

67,374

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

5 Answers5