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$1639)$ $$\int \frac{x^2}{ 1+x^2}\, \mathrm{d}x$$

I know that ,$$\int \frac{1}{ 1+x^2}\, \mathrm{d}x = \arctan x+C$$ but I dont know what to do with $x^2$, so I used integration by parts method, and got

$$x^2\arctan x + C - \int 2x\arctan x \mathrm{d}x$$

so what I really want, is to know,should I go further and again integrate by part method what I got, or there some elementary integral which I can use.

And I need some hint to solve this problem:
$$\int \frac{x^{2}+3}{x^2-1}\, \mathrm{d}x$$

I hope I explain my point clear,thanks.

  • 1
    Hint: polynomial long division. – 2017-01-07

3 Answers 3

3

Hint $1$: $$I_1 =\int dx -\int \frac {1}{1+x^2} dx $$

Hint $2$: $$I_2 =\int dx +\int \frac {4}{x^2-1 } dx $$

Hope you can take it from here.

3

Hint \begin{eqnarray} \dfrac{x^2}{1+x^2}&=&\dfrac{1+x^2-1}{1+x^2}=\dfrac{1+x^2}{1+x^2}-\dfrac{1}{1+x^2}=1-\dfrac{1}{1+x^2}\\ \dfrac{x^2+3}{1-x^2}&=&\dfrac{x^2-1+4}{1-x^2}=\dfrac{x^2-1}{1-x^2}+\dfrac{4}{1-x^2}=-1+2\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right) \end{eqnarray}

2

hint: $\dfrac{x^2}{1+x^2} = 1 - \dfrac{1}{1+x^2}$