Good integration question to try (1 Viewer)

J-Wang · Sep 11, 2012

$\int \sqrt{x^2+x+1}$
Try this. Its looks are deceiving

SpiralFlex · Sep 11, 2012

Not too deceptive but messy, posting up sol.

barbernator · Sep 11, 2012

iirc wanga, it was x^2-x but same way around it anyway. its a very good question

SpiralFlex · Sep 11, 2012

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

$$Let$\; x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan \theta$

$dx=\frac{\sqrt{3}}{2}\sec^2 \theta \;d \theta$

$=\int \sqrt{\frac{3}{4}(\tan^2 \theta +1)}*\frac{\sqrt{3}}{2}\sec^2 \theta \;d \theta$

$=\frac{3}{4} \int \sec^3 \theta \; d\theta$

$Use IBP,$

$=\frac{3}{4}*\frac{1}{2}(\sec x\tan x+ \ln (\sec x +\tan x))+C$

$Constructing a triangle,$

$$Hypotenuse:$\; 2\sqrt{x^2+x+1}$

$$Other sides:$\; 4x+2, 3$

$=\frac{3}{8}(\sec x \tan x+\ln (\sec x+ \tan x))+C$

$=\frac{1}{4}*\sqrt{x^2+x+1}*(2x+1)+\frac{3}{8}*\ln (\frac{2\sqrt{x^2+x+1}+(2x+1)}{\sqrt{3}})+C$

$Using the inverse hyperbolic identity.$

$\sinh^{-1}x=\ln (x+\sqrt{x^2+1})$

$=(\frac{x}{2}+\frac{1}{4})*\sqrt{x^2+x+1}+\frac{3}{8}\sinh^{-1} \frac{2x+1}{\sqrt{3}}+C$

Shadowdude · Sep 11, 2012

J-Wang said:
$\int \sqrt{x^2+x+1}$
Try this. Its looks are deceiving

Assuming integration is with respect to y, answer is:

$y\sqrt{x^2+x+1} + C$

Sindivyn · Sep 11, 2012

Shadowdude said:
Assuming integration is with respect to y, answer is:

$y\sqrt{x^2+x+1} + C$

lol'd

Shadowdude · Sep 11, 2012

barbernator said:
pls no. actually pls demonstrate your mathematical prowess by posting a solution.

tbh, if that popped up in my exam - i'd do it that way

it's not wrong.

Plus, my mind currently is on vector spaces and T-invariant eigensomethings... not on integration, unfortunately. I need to get into my integration zone.

SpiralFlex · Sep 11, 2012

Shadowdude said:
tbh, if that popped up in my exam - i'd do it that way

it's not wrong.

Plus, my mind currently is on vector spaces and T-invariant eigensomethings... not on integration, unfortunately. I need to get into my integration zone.

You have to leave the friend zone first.

Frostbitten · Sep 11, 2012

SpiralFlex said:
You have to leave the friend zone first.

Oh! Oh heeell no!

hayabusaboston · Sep 11, 2012

OMG i just died! LOL

deswa1 · Sep 11, 2012

SpiralFlex said:
You have to leave the friend zone first.

Nice.

2ndt0m3h · Sep 12, 2012

well 1+x+x^2 is the same as ((x^3)-1)/(x-1)

then let u=x-1 and finish off easy.

Nooblet94 · Sep 12, 2012

SpiralFlex said:
You have to leave the friend zone first.

Oh god, I just about died.

lolcakes52 · Sep 12, 2012

Cue whipping noise.

4025808 · Sep 12, 2012

I would have flopped out if that question appeared in my exam =_=

Good integration question to try (1 Viewer)

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