1.A particle that moves along a straight line has velocity meters per second after t seconds. How many meters will it travel during the first t seconds?
I used integration by parts but i didnt get the write answer.
2.
(1) You're going to need to use integration by parts twice.
First round requires: $\displaystyle u = t^2 $ and $\displaystyle dv = e^{-2t} dt$
You will get: $\displaystyle \int t^2 e^{-2t} = -\frac{1}{2}t^2e^{-2t} + 2 {\color{red}\int te^{-2t}dt}$
Use integration by parts again on the red: $\displaystyle u = t$ and $\displaystyle dv = e^{-2t}dt$
Can you carry on from here?
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(2) Again, you'll need integration by parts: $\displaystyle u = \arctan (3x)$ and $\displaystyle dv = xdx$
You'll get: $\displaystyle \int x\arctan (3x) dx = \frac{x^2}{2}\arctan (3x) - \frac{3}{2} {\color{blue}\int \frac{x^2}{1+9x^2} dx}$
The blue integral should be easier to deal with now. You'll need long division or you can notice that: $\displaystyle \frac{x^2}{1+9x^2} \cdot \frac{9}{9} = \frac{9x^2 \ {\color{red}+ 1 - 1}}{9 (1+9x^2)} = \frac{1}{9} - \frac{1}{9} \left( \frac{1}{1+9x^2}\right)$
$\displaystyle \int {x\arctan \left( {3x} \right)} dx = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{1}{2}\int {\frac{{3{x^2}}}{{{{\left( {3x} \right)}^2} + 1}}dx} = $
$\displaystyle = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{1}{6}\int {\frac{{{{\left( {3x} \right)}^2} + 1 - 1}}{{{{\left( {3x} \right)}^2} + 1}}dx} = $
$\displaystyle = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{6}\int {\frac{{dx}}{{{{\left( {3x} \right)}^2} + 1}}} =$
$\displaystyle = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{{18}}\int {\frac{{d\left( {3x} \right)}}{{{{\left( {3x} \right)}^2} + 1}}} = $
$\displaystyle = \frac{{{x^2}}}{2}\arctan \left( {3x} \right) - \frac{x}{6} + \frac{1}{{18}}\arctan \left( {3x} \right) + C =$
$\displaystyle = \frac{1}{{18}}\left( {9{x^2} + 1} \right)\arctan \left( {3x} \right) - \frac{x}{6} + C.$