On the evaluation of the integral $\int_{0}^{1}\sqrt{x^2+1}\,dx$ The integral $\displaystyle \int_{0}^{1}\sqrt{x^2+1}\,dx$ can be evaluated with the standard technic of sub $u=\tan \theta$. However, in the book says evaluate the integral without trigonometric substitution.
I can't find a way. I applied the sub $u=1-x$ and it got a little messie.
I also tried to approach it geometrically, but again I had a problem because the graph is not a common shape thus saying that the integral is equal to the area of that shape.
For example if I had the integral $\displaystyle \int_0^1 \sqrt{1-x^2}\,dx$ then this equal to aread of the quadrateral etc. In the case of the other integral I don't see something like this.
If someone could give me a hint,that would be nice!!
 A: To avoid trigonometric substitutions, one can use the change of variable $$t=\sqrt{1+\frac1{x^2}},$$ which leads to $$t\geqslant\sqrt2,\qquad\sqrt{1+x^2}=xt,\qquad x^2=\frac1{t^2-1},\qquad t\,\mathrm dt=\frac{\mathrm dx}{x^3},$$ hence $$\int_0^1\sqrt{1+x^2}\,\mathrm dx=\int_{\sqrt2}^\infty(tx)(x^3t\,\mathrm dt)=\int_{\sqrt2}^\infty\frac{t^2}{(t^2-1)^2}\,\mathrm dt.$$
As a rational function of $t$, the function in the last integral has an explicit primitive, hence we are done.
A: $$I=\int\sqrt{x^2+1}dx=x\sqrt{x^2+1}-\int x \frac{1}{2\sqrt{x^2+1}}2xdx=x\sqrt{x^2+1}-\int \frac{x^2-1+1}{\sqrt{x^2+1}}dx=x\sqrt{x^2+1}-\int\sqrt{x^2+1}dx-\int \frac{1}{\sqrt{x^2+1}}dt=x\sqrt{x^2+1}-I-\ln|x+\sqrt{x^2+1}|,$$so$$I=\frac{1}{2}\left(x\sqrt{x^2+1}-\ln(x+\sqrt{x^2+1})\right)+C.$$
A: Hint. Substitute: $\sqrt{x^2+1}=x+t$. This transforms your algebraic integral into a purely rational one:
$$\begin{align}
\int_{0}^{1}\sqrt{x^2+1}\,\mathrm{d}x
&=\int_{1}^{\sqrt{2}-1}\frac{t+\frac{1}{t}}{2}\cdot\frac{(-1)(t^2+1)}{2t^2}\,\mathrm{d}t
\end{align}$$
A: Hint
Set $x=\sinh u$ then
$$\int_0^1\sqrt{1+x^2}dx=\int_{0}^{\ln(1+\sqrt 2)}\sqrt{1+\sinh^2(u)}\cosh(u)du=\int_0^{\ln(1+\sqrt 2)}\cosh^2(u)du$$
now integrate by part with $a(u)=\cosh u$ and $b'(u)=\cosh(u)$
