t-2-1-t-2-2-dx-

Question Number 93484 by mashallah last updated on 13/May/20

\int t^{2} / {(1 + t^{2})}^{2} dx =

Commented by prakash jain last updated on 13/May/20

function is in t so constant wrt x

\frac{t^{2}}{{(1 + t^{2})}^{2}} x + C

Commented by mathmax by abdo last updated on 13/May/20

I = \int \frac{t^{2}}{{(1 + t^{2})}^{2}} d t \Rightarrow I = \int \frac{t^{2} + 1 - 1}{(1 + t^{2})} d t = \int \frac{d t}{1 + t^{2}} - \int \frac{d t}{{(1 + t^{2})}^{2}}

w e h a v e \int \frac{d t}{1 + t^{2}} = a r c t a n (t) + c_{1}

\int \frac{d t}{{(1 + t^{2})}^{2}} =_{t = t a n x} \int \frac{(1 + {t a n}^{2} x) d x}{{(1 + {t a n}^{2} x)}^{2}} = \int {c o s}^{2} x d x

= \frac{1}{2} \int (1 + c o s (2 x)) d x = \frac{x}{2} + \frac{1}{4} s i n (2 x) + c_{2}

s i n (2 x) = \frac{2 t a n x}{1 + {t a n}^{2} x} = \frac{2 t}{1 + t^{2}} \Rightarrow \int \frac{d t}{{(1 + t^{2})}^{2}} = \frac{1}{2} a r c t a n (t) + \frac{t}{2 (1 + t^{2})} + c_{2}

\Rightarrow I = \frac{1}{2} a r c t a n (t) - \frac{t}{2 (1 + t^{2})} + C