How do you integrate by substitution $\int t^{2} (t - \frac{2}{t}) d t$ $int t^2(t-2/t)dt$ ?

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How do you integrate by substitution $\int t^{2} (t - \frac{2}{t}) d t$ $int t^2(t-2/t)dt$ ?

1 Answer

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Answer 1 · 2017-05-28T03:57:15

$\frac{{(t^{2} - 2)}^{2}}{4} + C$ $((t^2-2)^2)/4+C$

Explanation:

$\int t^{2} (t - \frac{2}{t}) d t$ $intt^2(t-2/t)dt$
$\int t (t^{2} - 2) d t$ $intt(t^2-2)dt$

Substitute $u = t^{2} - 2$ $u=t^2-2$ , $\frac{d u}{d x} = 2 t$ $(du)/(dx) =2t$

$\frac{1}{2} \int 2 t (t^{2} - 2) d t$ $1/2int2t(t^2-2)dt$
$\frac{1}{2} \int (u) d u$ $1/2int(u)du$
$\frac{1}{2} (\frac{u^{2}}{2})$ $1/2((u^2)/2)$
$\frac{u^{2}}{4}$ $(u^2)/4$

Undo substitution:
$\frac{{(t^{2} - 2)}^{2}}{4} + C$ $((t^2-2)^2)/4+C$

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How do you integrate by substitution $\int t^{2} (t - \frac{2}{t}) d t$ $int t^2(t-2/t)dt$ ?

1 Answer

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How do you integrate by substitution $\int t^{2} (t - \frac{2}{t}) d t$ $int t^2(t-2/t)dt$ ?