So is it safe to say that we can use integration by substitution???

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- Dec 13th 2012, 09:11 PMasilvester635Integration help
So is it safe to say that we can use integration by substitution???

- Dec 13th 2012, 09:14 PMMarkFLRe: Integration help
I would simplify the integrand by factoring $\displaystyle \sqrt{t}$ from the numerator, then dividing out the common factors of the numerator and denominator. Or you might find it easier to just split the fraction, then reduce each term.

- Dec 13th 2012, 09:14 PMbkbowserRe: Integration help
carry out the division

- Dec 13th 2012, 09:36 PMasilvester635Re: Integration help
Can' i just put √t on the numerator and make it t^-1/2 and multiply (t + 2t^2)t^-1/2 together and integrate????

- Dec 13th 2012, 09:41 PMasilvester635Re: Integration help
Can' i just put √t on the numerator and make it t^-1/2 and multiply (t + 2t^2)t^-1/2 together and integrate????.....

- Dec 13th 2012, 09:46 PMMarkFLRe: Integration help
Yes, that's another way of splitting the fraction.

- Dec 13th 2012, 09:47 PMbkbowserRe: Integration help
yup.

you'd end up with $\displaystyle \frac{t}{t^{1/2}}+\frac{2t^2}{t^{1/2}}=t^{1/2}+2*t^{3/2}$

or $\displaystyle \int t^{1/2}+2*t^{3/2}dx$ - Dec 13th 2012, 10:09 PMasilvester635Re: Integration help
then im authorized to integrate after that step right?

- Dec 13th 2012, 10:13 PMMarkFLRe: Integration help
Yes, now it is just a matter of applying the power rule term by term.

- Dec 17th 2012, 08:07 PMx3bnmRe: Integration help
You can do this with the help of integration by parts.

Now after simplification:

$\displaystyle \int \frac{t + 2t^2}{\sqrt{t}}\;\; dt = \int (\sqrt{t} + 2t\sqrt{t})\;\;dt = \int \sqrt{t}\;\;dt + \int 2t\sqrt{t}\;\;dt$

We divide this integration in two parts. First:

$\displaystyle \int \sqrt{t}\;\;dt = \frac{2}{3}t^{\frac{3}{2}} + C_1$

Then for $\displaystyle \int 2t\sqrt{t}\;\; dt$ we use integration by parts:

$\displaystyle \text{Let } u = t \;\;\therefore du = dt$

$\displaystyle \text{Let } dv = 2\sqrt{t}\;\; dt\;\; \therefore v = \frac{4}{3}t^{\frac{3}{2}}$

Now:

$\displaystyle \begin{align*}\int 2t\sqrt{t}\;\;dt =& \frac{4}{3}t^{\frac{5}{2}} - \int \frac{4}{3}t^{\frac{3}{2}}\;\; dt \\ =& \frac{4}{3}t^{\frac{5}{2}} - \frac{8}{15}t^{\frac{5}{2}} + C_2 \\ =& \frac{4}{5}t^{\frac{5}{2}}+ C_2...........\text{[After simplicfication]}\end{align*}$

So:

$\displaystyle \therefore \int \frac{t + 2t^2}{\sqrt{t}}\;\; dt = \frac{2}{3}t^{\frac{3}{2}} + \frac{4}{5}t^{\frac{5}{2}} + C$

Check:

integration (t + 2*t^2)/(t)^(1/2) dt - Wolfram|Alpha