1. difficult integral

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

2. 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

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

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

4. 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

5. Thank you very much

Regards