Question

Integral of x^2/(x^2+2)

Answer 1

\[\int\limits X^2dx/(X^2+2)\]

Answer 2

\[\int\limits \frac{ x^2}{ X^2+2 }\]

Answer 3

first do long division

Answer 4

how can you tell that?

Answer 5

I will, one sec

Answer 6

because of similarity?

Answer 7

\[\int\limits 1-\frac{ 2 }{ x^2+1 }\]

Answer 8

because the exponent of the numerator is the same (or larger than) the denominator, we should do long division

Answer 9

ok

Answer 10

from here I didn't get the same answer as written though...

Answer 11

I got x-arctanX

Answer 12

$$

\int\limits 1-\frac{ 2 }{ x^2+\color{red}2 }

$$

Answer 13

yeah. and from here I divide it into two integrals

Answer 14

$$
\Large{
\int\limits \frac{ x^2}{ x^2+2 }
=\int\limits 1-\frac{ 2 }{ x^2+2 }\\
=x - 2\cdot \int \frac{1}{x^2+2}\\
=x - 2\cdot \int \frac{1}{2(\frac{x^2}{2}+1)}\\
=x - 2\cdot \int \frac{1}{2~( (\frac{x}{\sqrt{2} })^2+1)}\\
=x - \frac{2}{2}\cdot \int \frac{1}{ (\frac{x}{\sqrt{2} })^2+1)}\\
=x - \int \frac{1}{ (\frac{x}{\sqrt{2} })^2+1)}\\
\\~~\\
u = \frac{x}{\sqrt{2}}\\
du = ...
}
$$
Now apply u substitution

Answer 15

you're trully brilliant !!! I've got it! Thanks allot helping me out!!!!!!!!

Answer 16

for*

Answer 17

Directly substituting a trig expression works nicely as well. Let \(x=\sqrt2\tan t\), then \(dx=\sqrt2\sec^2t\,dt\), then
\[\int\frac{x^2}{x^2+2}\,dx=\int\frac{\sqrt2\sec^2t}{2\tan^2t+2}\,dt=\frac{1}{\sqrt2}\int\,dt=\cdots\]