# problematic integral #2

• Mar 21st 2010, 02:12 AM
Shock
problematic integral #2
Hi.
Evaluate:

$\displaystyle \int {\sqrt{x^2+x+3} dx}$
• Mar 21st 2010, 02:15 AM
mr fantastic
Quote:

Originally Posted by Shock
Hi.
Evaluate:

$\displaystyle \int {\sqrt{x^2+x+3} dx}$

See here: integrate Sqrt&#x5b;x&#x5e;2 &#x2b; x &#x2b; 3&#x5d; - Wolfram|Alpha

(Click on Show steps).
• Mar 21st 2010, 02:54 AM
Moo
Hello,

By completing the square, we have $\displaystyle x^2+x+3=\left(x+\tfrac 12\right)^2+\tfrac{11}{4}=\tfrac{11}{4}\left(\tfra c{4}{11}\cdot\left(x+\tfrac 12\right)^2+1\right)$

Using the usual substitutions, let $\displaystyle \sinh(u)=\sqrt{\tfrac{4}{11}}\cdot\left(x+\tfrac 12\right)$

(thus $\displaystyle \cosh(u) ~du=\sqrt{\tfrac{4}{11}} ~dx$)

And you'll have $\displaystyle \int \sqrt{\tfrac{11}{4}}\cdot\sqrt{\tfrac{11}{4}} \cdot \sqrt{\sinh^2(u)+1}\cdot\cosh(u) ~du=\tfrac{11}{4}\int \cosh^2(u) ~du$

But $\displaystyle 2\cosh^2(u)=1+\cosh(2u)$

So $\displaystyle \tfrac{11}{4}\int \cosh^2(u) ~du=\tfrac{11}{8}\left(x+\tfrac 12 \cdot \sinh(2u)\right)+C$

Now, back substitute :D