difficult integral

**chocolatelover** · August 28th 2007, 10:21 AM

Hi everyone,

Can you tell me if this is correct, please?

Evaluate the following integral:

x/(1+x^4 )dx

u=x/(1+x^4)
du=arctanx dx

x/1+x^4=
udu=
(x/(1+x^4))arctanx+c

Thank you very much

**Jhevon** · August 28th 2007, 10:33 AM

Originally Posted by chocolatelover

Hi everyone,

Can you tell me if this is correct, please?

Evaluate the following integral:

x/(1+x^4 )dx

u=x/(1+x^4)
du=arctanx dx

x/1+x^4=
udu=
(x/(1+x^4))arctanx+c

Thank you very much

i'm afraid that's incorrect. here's one way:

$\int \frac {x}{1 + x^4}~dx = \int \frac {x}{1 + \left( x^2 \right)^2}~dx$

We proceed by substitution

Let $u = x^2$

$\Rightarrow du = 2x~dx$

$\Rightarrow \frac {1}{2} du = x~dx$

So our integral becomes:

$\frac {1}{2} \int \frac {1}{1 + u^2}~du$

$= \frac {1}{2} \arctan u + C$

$= \frac {1}{2} \arctan \left(x^2 \right) + C$

Allow me to stress that $\int \frac {1}{1 + x^2}~dx = \arctan x + C$ NOT $\frac {d}{dx} \left( \frac {1}{1 + x^2} \right) = \arctan x$

and you didn't even have that, you had $\frac {d}{dx} \left( \frac {x}{1 + x^4}\right) = \arctan x$ now that's just wrong

**Krizalid** · August 28th 2007, 03:04 PM

$\int {\frac{x}<br /> {{1 + x^4 }}\,dx} = \frac{1}<br /> {2}\int {\frac{{\left( {x^2 } \right)'}}<br /> {{1 + \left( {x^2 } \right)^2 }}\,dx} = \frac{1}<br /> {2}\arctan x^2 + k,\,k\in\mathbb R$

Jhevon, don't be so hard, not necessary to rub her(his) in, the "not"

**Jhevon** · August 28th 2007, 03:07 PM

Originally Posted by Krizalid

Jhevon, don't be so hard, not necessary to rub her(his) in, the "not"

it wasn't my intention to be harsh.

but i bet he/she will never forget now

it's just tough love

**chocolatelover** · August 28th 2007, 06:04 PM

Thank you very much

Regards

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