# How do you find the integral int t^2(t^3+4)^(-1/2)dt using substitution?

Mar 13, 2017

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

#### 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: $\int {t}^{2} {\left({t}^{3} + 4\right)}^{- \frac{1}{2}} \mathrm{dt}$, there is a ${t}^{3}$ and a ${t}^{2}$ which hints that we may want to incorporate ${t}^{3}$ into our substitution since its derivative is present.

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

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

$\frac{1}{3} \int \frac{1}{u} ^ \left(\frac{1}{2}\right) \mathrm{du} = \frac{1}{3} \int {u}^{- \frac{1}{2}} \mathrm{du} = \left(\frac{1}{3}\right) \left(2 {u}^{\frac{1}{2}}\right) + C = \frac{2}{3} {u}^{\frac{1}{2}} + C$

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

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

Mar 13, 2017

$\int {t}^{2} {\left({t}^{3} + 4\right)}^{- \frac{1}{2}} \mathrm{dt} = \frac{2}{3} {\left({t}^{3} + 4\right)}^{\frac{1}{2}} + C$

#### Explanation:

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

$\int {t}^{2} {\left({t}^{3} + 4\right)}^{- \frac{1}{2}} \mathrm{dt}$

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

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

differentiate this using the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \times 3 {t}^{2} {\left({t}^{3} + 4\right)}^{- \frac{1}{2}}$

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

$\therefore \int {t}^{2} {\left({t}^{3} + 4\right)}^{- \frac{1}{2}} \mathrm{dt} = \frac{2}{3} {\left({t}^{3} + 4\right)}^{\frac{1}{2}} + C$

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