1. ## calculus 2/integrals

Hi everyone,

Can you please tell me if this is correct and if it isn't can you give me a hint or explain to me why it is wrong?

Evaluate the following integrals:

X/(X^2+1)^2dx

u=x^2+1
du+2xdx
xdx=du/2

x/(x^2+1)^2dx=
1/u^2du/2=
1/lnu^2+c=
1/2ln(x^2+1)^2

Thank you very much

P.S.
Does anyone know how to add math symbols on the computor? that is, how to add a square root, x squared etc.

2. Originally Posted by chocolatelover
Hi everyone,

Can you please tell me if this is correct and if it isn't can you give me a hint or explain to me why it is wrong?

Evaluate the following integrals:

X/(X^2+1)^2dx

u=x^2+1
du+2xdx
xdx=du/2

x/(x^2+1)^2dx=
1/u^2du/2=
1/2ln(x^2+1)^2

...
Hello,

remember $\int \frac{1}{x^2}dx = -\frac{1}{x} +C$

3. Originally Posted by chocolatelover
Hi everyone,

Can you please tell me if this is correct and if it isn't can you give me a hint or explain to me why it is wrong?

Evaluate the following integrals:

X/(X^2+1)^2dx

u=x^2+1
du+2xdx
xdx=du/2

x/(x^2+1)^2dx=
1/u^2du/2=
1/lnu^2+c=
1/2ln(x^2+1)^2

Thank you very much
You started off nicely, then it went horribly wrong ... well, not horribly, i'm exagerated, i can see how you made that mistake

doing what you did:

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

We proceed by substitution:

Let $u = x^2 + 1$

$\Rightarrow du = 2x~dx$

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

So our integral becomes:

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

THIS IS WHERE YOU MADE THE MISTAKE, THIS INTEGRAL DOES NOT GIVE $\color {red} \ln (u)$

$\frac {1}{2} \int \frac {1}{u^2}~du = \frac {1}{2} \int u^{-2}~du$ .........now use the power rule

$= - \frac {1}{2}u^{-1} + C$

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

P.S.
Does anyone know how to add math symbols on the computor? that is, how to add a square root, x squared etc.
do you know how to use LaTex? see the LaTex Tutorial on this site

otherwise, to say "the square root of x" you would type "sqrt(x)"

4. Thank you very much

Regards