# Integration help

• Dec 13th 2012, 09:11 PM
asilvester635
Integration help
So is it safe to say that we can use integration by substitution???
• Dec 13th 2012, 09:14 PM
MarkFL
Re: Integration help
I would simplify the integrand by factoring $\sqrt{t}$ 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 PM
bkbowser
Re: Integration help
carry out the division
• Dec 13th 2012, 09:36 PM
asilvester635
Re: 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 PM
asilvester635
Re: 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 PM
MarkFL
Re: Integration help
Yes, that's another way of splitting the fraction.
• Dec 13th 2012, 09:47 PM
bkbowser
Re: Integration help
yup.

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

or $\int t^{1/2}+2*t^{3/2}dx$
• Dec 13th 2012, 10:09 PM
asilvester635
Re: Integration help
then im authorized to integrate after that step right?
• Dec 13th 2012, 10:13 PM
MarkFL
Re: Integration help
Yes, now it is just a matter of applying the power rule term by term.
• Dec 17th 2012, 08:07 PM
x3bnm
Re: Integration help
Quote:

Originally Posted by asilvester635
So is it safe to say that we can use integration by substitution???

You can do this with the help of integration by parts.

Now after simplification:

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

We divide this integration in two parts. First:

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

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

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

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

Now:

\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:

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

Check:

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