# How can I find te integral of root(x*x+x+1)

1. Jan 2, 2005

### Tengo

How can I find te integral of root(x*x+x+1)Thank you

2. Jan 2, 2005

### HallsofIvy

Staff Emeritus
x2+ x+ 1= x2+ x+ 1/4+ 3/4= (x+ 1/2)2+ 3/4.

Let u= x+ 1/2.

3. Jan 2, 2005

### Pyrrhus

You need to put this integral in the form of

$$\int \sqrt{a^2 + u^2} du$$

This can be achieved by completing the square.

4. Jan 2, 2005

### dextercioby

$$I=\int \sqrt{x^{2}+x+1} dx =\int \sqrt{(x+\frac{1}{2})^{2}+\frac{3}{4}} dx$$ (1)

Like Halls said,make the first substitution $x+\frac{1}{2}\rightarrow u$ (2)
and u'll get
$$I=\int \sqrt{u^{2}+\frac{3}{4}} du$$ (3)
Write this integral as
$$I=\int \sqrt{\frac{3}{4}} \sqrt{(\sqrt{\frac{4}{3}} u)^{2} +1} du$$ (4)
and make the substitution
$$\sqrt{\frac{4}{3}} u\rightarrow \sinh v$$ (5)
,under which:
$$du\rightarrow \sqrt{\frac{3}{4}} \cosh v dv$$ (6)
,so the integral becomes:
$$I=\frac{3}{4}\int \cosh^{2}v dv$$ (7)
,where i have made use of the fundamental formula of hyperbolic trigometry
$$\cosh^{2}v-\sinh^{2}v =1$$ (8)
Use the formula of the double angle:
$$\cosh^{2}v=\frac{1+\cosh 2v}{2}$$(9)
to solve the integral in terms of 'v'.Return to the initial variable 'x' via the substitutions (5) and (2).

Daniel.