Find the indefinite integral

**ugkwan** · November 15th 2010, 12:28 AM

Hi,

$\displaystyle\int\frac{4}{4x^2+4x+65}dx$

Hints please. I am stumped by this problem. Maybe completing the square?

**Prove It** · November 15th 2010, 12:36 AM

Yes, complete the square...

$\displaystyle \int{\frac{4}{4x^2 + 4x + 65}\,dx} = \int{\frac{1}{x^2 + x + \frac{65}{4}}\,dx}$

$\displaystyle = \int{\frac{1}{x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + \frac{65}{4}}\,dx}$

$\displaystyle = \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{65}{4}}\,dx}$

$\displaystyle = \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 + 16}\,dx}$ .

Now make the substitution $\displaystyle x + \frac{1}{2} = 4\sinh{t}$ so that $\displaystyle dx = 4\cosh{t}\,dt$ and the integral becomes

$\displaystyle \int{\frac{1}{(4\sinh{t})^2 + 16}\,4\cosh{t}\,dt}$

$\displaystyle = \int{\frac{4\cosh{t}}{16(\sinh^2{t} + 1)}\,dt}$

$\displaystyle = \int{\frac{\cosh{t}}{4\cosh{t}}\,dt}$

$\displaystyle = \int{\frac{1}{4}\,dt}$

$\displaystyle = \frac{t}{4} + C$

$\displaystyle = \frac{\sinh^{-1}\left(\frac{x}{4} + \frac{1}{8}\right)}{4} + C$ .

**FernandoRevilla** · November 15th 2010, 12:40 AM

Originally Posted by ugkwan

Maybe completing the square?

Yes,

$4x^4+4x+65=4(x+\;1/2)^2+64$

Now use $t=2(x+\;1/2)$ , etc.

Regards.

Sorry Prove It, I didn't see your answer.

**HallsofIvy** · November 15th 2010, 03:19 AM

Or: from $\displaystyle \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 + 16}\,dx}$
let 4u= x+ 1/2, so that 4du= dx and the integral becomes
$\displaystyle \int{\frac{1}{\left(16u^2 + 16}\, 4du}= \frac{1}{4}\int\frac{1}{u^+ 1}du$
$\displaystyle = \frac{1}{4}arctan(u)+ C$
$\displaystyle= \frac{1}{4}arctan\left(\frac{1}{4}(x+ 1/2)\right)+ C$

**tonio** · November 15th 2010, 03:20 AM

Originally Posted by HallsofIvy

Or: from $\displaystyle{ \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 + 16}\,dx}$
let 4u= x+ 1/2, so that 4du= dx and the integral becomes
$\displaystyle \int{\frac{1}{\left(16u^2 + 16}\, 4du}= \frac{1}{4}\int\frac{1}{u^+ 1}du$
$\displaystyle = \frac{1}{4}arctan(u)+ C$

.

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