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


*

*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 $\tan$ ?
 A: 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,
$$I=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}u=\frac{1}{\sqrt{2}}.$$
A: These are some kind of standard trigonometric substitutions:
$$\sqrt{x^2-a^2}\text{ or }x^2-a^2\tag{$x=a\sin\theta$ or $x=a\cos\theta$}$$
$$\sqrt{x^2+a^2}\text{ or }x^2+a^2\tag{$x=a\tan\theta$ or $x=a\cot\theta$}$$
A: 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.
A: 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.
A: 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 $$\int(-1/t^2)dt$$
