How do you integrate $\int t^{2} {(t^{3} + 4)}^{- \frac{1}{2}}$ $int t^2(t^3+4)^(-1/2)$ using substitution?

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Answer 1 · 2018-04-03T22:02:23

The integral is $\frac{2}{3} \sqrt{t^{3} + 4} + C$ $2/3sqrt(t^3+4)+C$

First, rewrite the integral:

$= \int t^{2} {(t^{3} + 4)}^{- \frac{1}{2}} d t$ $color(white)=int t^2(t^3+4)^(-1/2)dt$

$= \int \frac{t^{2}}{{(t^{3} + 4)}^{\frac{1}{2}}} d t$ $=int t^2/(t^3+4)^(1/2)dt$

$= \int \frac{t^{2}}{\sqrt{t^{3} + 4}} d t$ $=int t^2/sqrt(t^3+4)dt$

Now, let:

$u=t^3+4quadcolor(blue)=>quaddu=3t^2dtquadcolor(blue)=>quaddt=(du)/(3t^2)$

Substituting:

$=int t^2/sqrt(u)*(du)/(3t^2)$

$=int color(red)cancelcolor(black)(t^2)/sqrt(u)*(du)/(3color(reD)cancelcolor(black)(t^2))$

$=int 1/sqrt(u)*(du)/3$

$=1/3int1/sqrtudu$

$=1/3int1/u^(1/2)du$

$=1/3intu^(-1/2)du$

$=1/3*u^(-1/2+1)/(-1/2+1)$

$=1/3*u^(1/2)/(1/2)$

$=1/3*2*sqrtu$

$=2/3sqrtu$

Put $t^3+4$ back in for $u$ (and don't forget to add $C$ ):

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

That's it. Hope this helped!

How do you integrate int t^2(t^3+4)^(-1/2)∫t2(t3+4)−12int t^2(t^3+4)^(-1/2) using substitution?