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:
$\displaystyle \int \frac {x}{1 + x^4}~dx = \int \frac {x}{1 + \left( x^2 \right)^2}~dx$
We proceed by substitution
Let $\displaystyle u = x^2$
$\displaystyle \Rightarrow du = 2x~dx$
$\displaystyle \Rightarrow \frac {1}{2} du = x~dx$
So our integral becomes:
$\displaystyle \frac {1}{2} \int \frac {1}{1 + u^2}~du$
$\displaystyle = \frac {1}{2} \arctan u + C$
$\displaystyle = \frac {1}{2} \arctan \left(x^2 \right) + C$
Allow me to stress that $\displaystyle \int \frac {1}{1 + x^2}~dx = \arctan x + C$ NOT $\displaystyle \frac {d}{dx} \left( \frac {1}{1 + x^2} \right) = \arctan x $
and you didn't even have that, you had $\displaystyle \frac {d}{dx} \left( \frac {x}{1 + x^4}\right) = \arctan x$ now that's just wrong
$\displaystyle \int {\frac{x}
{{1 + x^4 }}\,dx} = \frac{1}
{2}\int {\frac{{\left( {x^2 } \right)'}}
{{1 + \left( {x^2 } \right)^2 }}\,dx} = \frac{1}
{2}\arctan x^2 + k,\,k\in\mathbb R$
Jhevon, don't be so hard, not necessary to rub her(his) in, the "not"