How do you find the integral $\int t^{2} {(t^{3} + 4)}^{- \frac{1}{2}} d t$ $int t^2(t^3+4)^(-1/2)dt$ using substitution?

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How do you find the integral $\int t^{2} {(t^{3} + 4)}^{- \frac{1}{2}} d t$ $int t^2(t^3+4)^(-1/2)dt$ using substitution?

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Answer 1 · 2017-03-13T02:26:22

$\frac{2}{3} {(t^{3} + 4)}^{\frac{1}{2}} + C$ $2/3(t^3+4)^(1/2)+C$ or $\frac{2}{3} \sqrt{t^{3} + 4} + C$ $2/3sqrt(t^3+4)+C$

Explanation:

The key to solving any integral is to see what "type" of integral it could classify as. When I see an integral I try to ask myself if it is a substitution, integration by parts, trigonometric, trig sub, or partial fractions integral. In order to know whether or not to use the substitution method is whether or not the integral has both a function and its derivative present in the integral.

For this integral: $\int t^{2} {(t^{3} + 4)}^{- \frac{1}{2}} d t$ $intt^2(t^3+4)^(-1/2)dt$ , there is a $t^{3}$ $t^3$ and a $t^{2}$ $t^2$ which hints that we may want to incorporate $t^{3}$ $t^3$ into our substitution since its derivative is present.

Let's do just that.
$\int \frac{t^{2}}{{(t^{3} + 4)}^{\frac{1}{2}}} d t$ $intt^2/(t^3+4)^(1/2)dt$

Let $u = t^{3} + 4$ $u=t^3+4$
so, $d u = 3 t^{2} d t$ $du=3t^2dt$ or $\frac{1}{3} d u = t^{2} d t$ $1/3du=t^2dt$

$\frac{1}{3} \int \frac{1}{u^{\frac{1}{2}}} d u = \frac{1}{3} \int u^{- \frac{1}{2}} d u = (\frac{1}{3}) (2 u^{\frac{1}{2}}) + C = \frac{2}{3} u^{\frac{1}{2}} + C$ $1/3int1/u^(1/2)du=1/3intu^(-1/2)du=(1/3)(2u^(1/2))+C=2/3u^(1/2)+C$

Putting $t$ $t$ back in, we get:
$\frac{2}{3} {(t^{3} + 4)}^{\frac{1}{2}} + C$ $2/3(t^3+4)^(1/2)+C$

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

Answer 2 · 2017-03-13T08:39:43

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

Explanation:

Because the function outside the bracket is a constant $\times$ $xx$ the derivative of the bracket , it suggests that we might be able to do this by 'inspection'.

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

using the power rule for integration let us 'guess' that teh integral is of teh form

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

differentiate this using the chain rule:

$\frac{d y}{d x} = \frac{1}{2} \times 3 t^{2} {(t^{3} + 4)}^{- \frac{1}{2}}$ $(dy)/(dx)=1/2xx3t^2(t^3+4)^(-1/2)$

comparing this with the original question we see that the only difference is the constant $\frac{3}{2}$ $3/2$ so we adjust our 'guess' by a suitable value.

$:.intt^2(t^3+4)^(-1/2)dt=2/3(t^3+4)^(1/2)+C$

If one can spot integrals by inspection it can save a lot of work!.

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How do you find the integral $\int t^{2} {(t^{3} + 4)}^{- \frac{1}{2}} d t$ $int t^2(t^3+4)^(-1/2)dt$ using substitution?

2 Answers

Explanation:

Explanation:

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How do you find the integral int t^2(t^3+4)^(-1/2)dt∫t2(t3+4)−12dtint t^2(t^3+4)^(-1/2)dt using substitution?

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Explanation:

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How do you find the integral $\int t^{2} {(t^{3} + 4)}^{- \frac{1}{2}} d t$ $int t^2(t^3+4)^(-1/2)dt$ using substitution?