Indefinite Integral Problem

**njdurkin** · May 1st 2012, 06:41 AM

*SOLVED*
Calculus Single Variable 5th Edition
pg 310 problem #51

I'm having trouble with this one problem:

Find the indefinite integral:

Indefinite Integral Problem-math_image22.gif

The book's answer is: (edit-this was the correct answer)

Indefinite Integral Problem-math_image3.gif

Mathway gives an answer of:

Indefinite Integral Problem-math_image4.gif

I don't know how to even start this problem besides taking the integral of t x square root of t and adding that to the integral of the second part. Just simple indefinite integral rule. But when they are multiplied I'm not sure what to do. Substitution hasn't been covered yet so I'm assuming you don't use substitution. It just bothers me that I can't do this one problem and I got 25/26 assigned correct but this 26th one I can't get.
*SOLVED*

**TheIntegrator** · May 1st 2012, 07:03 AM

It might help to separate the integral into a couple pieces :

$\int t \sqrt(t) dt = \int t*t^{1/2} dt = \int t^{3/2} dt = \frac{t^{3/2+1}}{(3/2+1)} = \frac{2}{5} t^{5/2}$

$\int \frac{1}{(\sqrt t)} dt = \int t^{-1/2} dt = \frac{t^{-1/2+1}}{(-1/2+1)} = 2 t^{1/2} = -2 \sqrt(t)$

$\int (t \sqrt(t) + \frac{1}{(\sqrt t)} dt = \frac{2}{5} t^{5/2} - 2 \sqrt(t) + C$

do you see now? It helps when you see roots to express them as an exponent and then for this problem use the general power rule

**njdurkin** · May 1st 2012, 07:18 AM

Ah so simple! I can't believe I couldn't figure that out! It was so easy!

Thanks! (:

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