1. ## Integrate (2+3x^2)/(x^2(1+x^2))

$\int \frac {2+3x^2}{x^2(1+x^2)}dx$

I tried to substitute $1+x^2$ as 't'. But that failed.
Then i divided all terms by $x^2$, and that didnt give me any result too..

Finally, when i saw the answer at the back, i realized that i could write this as a sum of 2 terms..

But is there any easy method to do this? Without going through all this..?

2. ## Re: Integrate (2+3x^2)/(x^2(1+x^2))

Hello, animesh271094!

I found two ways . . . Neither is "easy".

$\int \frac {2+3x^2}{x^2(1+x^2)}dx$

Partial Fractions

$\frac{3x^2+2}{x^2(x^2+1)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}$

We find that: . $A = 0,\;B = 2,\;C = 0,\;D = 1$

We have: . $\int\left(\frac{2}{x^2} + \frac{1}{x^2+1}\right)\,dx \;=\; -\frac{2}{x} + \tan^{\text{-}1}\!x + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Substitution

Let $x = \tan\theta \quad\Rightarrow\quad dx = \sec^2\theta\,d\theta$

We have: . $\int\frac{2 + 3\tan^2\!\theta}{\tan^2\!\theta\,\sec^2\!\theta}\, \sec^2\!\theta\,d\theta \;=\;\int\frac{2+3\tan^2\!\theta}{\tan^2\!\theta}d \theta$

. . $=\;\int\left(2\cot^2\!\theta + 3\right)\,d\theta \;=\; \int\left[2(\csc^2\!\theta - 1) + 3\right]\,d\theta$

. . $=\;\int\left(2\csc^2\!\theta + 1\right)\,d\theta \;=\; -2\cot\theta + \theta + C$

Back-substitute: . $-2\left(\frac{1}{x}\right) + \tan^{\text{-}1}\!x + C \;=\;-\frac{2}{x} + \tan^{\text{-}1}\!x + C$

3. ## Re: Integrate (2+3x^2)/(x^2(1+x^2))

hi animesh271094

did you try using partial fractions?

$\frac{2+3x^2}{x^2(1+x^2)}=\frac{A}{x}+\frac{B}{x^2 }+\frac{Cx+D}{1+x^2}$

just find A,B,c and D.then it will be easier to integrate.

4. ## Re: Integrate (2+3x^2)/(x^2(1+x^2))

Thanx a lot guys..

5. ## Re: Integrate (2+3x^2)/(x^2(1+x^2))

There is an easier decomposition.
$\frac{2+3x^2}{x^2(1+x^2)}=\frac{2}{x^2}+\frac{1}{1 +x^2}$

6. ## Re: Integrate (2+3x^2)/(x^2(1+x^2))

hi Plato

that is the same as Soroban's decomposition.

7. ## Re: Integrate (2+3x^2)/(x^2(1+x^2))

Originally Posted by animesh271094
$\int \frac {2+3x^2}{x^2(1+x^2)}dx$
I always avoid partial fractions anytime I can.

We have $\frac{2+3{{x}^{2}}}{{{x}^{2}}\left( 1+{{x}^{2}} \right)}=\frac{2+2{{x}^{2}}+{{x}^{2}}}{{{x}^{2}}( 1+{{x}^{2}})}=\frac{2\left( 1+{{x}^{2}} \right)+{{x}^{2}}}{{{x}^{2}}\left( 1+{{x}^{2}} \right)}=\frac{2}{{{x}^{2}}}+\frac{1}{1+{{x}^{2}}} .$