How do you integrate by substitution $int (t+2t^2)/sqrtt dt$ ?

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Answer 1 · 2017-12-26T23:06:10

$int(t+2t^2)/sqrt(t)dt=2sqrt(t^3)/3+4sqrt(t^5)/5+C$

We see a $sqrt(t)$ our first thought is to substitude $u=sqrt(t)$ .

So let's do that :

$u=sqrt(t)=>du=1/(2sqrt(t))dt=>dt=2sqrt(t)du=>dt=2udu$

So our integral becomes :

$int(t+2t^2)/sqrt(t)dt=int(u^2+2u^4)/u*2udu=2int(u^2+2u^4)du=$

$2(u^3/3+2u^5/5)+C=2u^3/3+4u^5/5+C$

Now let's substitude $u=sqrt(t)$ back in :

$int(t+2t^2)/sqrt(t)dt=2sqrt(t^3)/3+4sqrt(t^5)/5+C$

How do you integrate by substitution int (t+2t^2)/sqrtt dtint (t+2t^2)/sqrtt dt?