How do you integrate $int troot3(t-4)dt$ ?

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Answer 1 · 2017-01-29T23:47:30

$3/7(t-4)^(7/3)+3(t-4)^(4/3)+C$

$I=inttroot3(t-4)color(white).dt$

Apply the substitution $u=t-4$ . This also implies that $t=u+4$ and $du=dt$ . Then:

$I=int(u+4)root3ucolor(white).du$

We can write $root3u$ with a fractional exponent and then distribute:

$I=int(u+4)u^(1/3)color(white).du$

$I=intu^(4/3)color(white).du+4intu^(1/3)color(white).du$

Integrate both using the rule $intu^ncolor(white).du=u^(n+1)/(n+1)+C$ :

$I=u^(7/3)/(7/3)+4(u^(4/3)/(4/3))+C$

$I=3/7u^(7/3)+3u^(4/3)+C$

Since $u=t-4$ :

$I=3/7(t-4)^(7/3)+3(t-4)^(4/3)+C$

How do you integrate int troot3(t-4)dtint troot3(t-4)dt?