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Integration by parts

$$x\tan^{-1}x~dx=\frac12\left(x^2+1\right)\tan^{-1}x-\frac12x+c$$

and

$$x^2e^{-3x}dx =-\frac13e^{-3x}\left(x^2+\frac23x+\frac29\right)+c$$

I have

$$d(uv)=u~dv+v~du$$

Can anyone please help me?

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I am a bit confused as to your question. Do you wish to see how these integrals are solved using integration by parts? –  Kris Williams May 12 '13 at 2:17
    
Please check to be sure that I interpreted everything correctly. Would you like me to add the missing integral signs? –  Brian M. Scott May 12 '13 at 2:17

2 Answers 2

up vote 1 down vote accepted

First one:

Recall that $d(\arctan(x)) = \dfrac{dx}{1+x^2}$. $$\int x \arctan(x) dx = \underbrace{\int \arctan(x) d\left(\dfrac{x^2}2 \right)}_{\displaystyle \int v du} = \overbrace{\dfrac{x^2 \arctan(x)}2}^{\displaystyle uv} - \underbrace{\int \dfrac{x^2}{2(1+x^2)}dx}_{\displaystyle \int u dv}$$ Now $$\int \dfrac{x^2}{2(1+x^2)}dx = \int \dfrac{1+x^2-1}{2(1+x^2)}dx = \int \dfrac{dx}2 - \int \dfrac{dx}{2(1+x^2)} = \dfrac{x}2 - \dfrac{\arctan(x)}2$$ Hence, we get that $$\int x \arctan(x) dx = \dfrac{x^2 \arctan(x)}2 - \dfrac{x}2 + \dfrac{\arctan(x)}2 + c = \dfrac{(1+x^2) \arctan(x)}2 - \dfrac{x}2 + c$$


Second one:

$$\int x^2 e^{-3x} dx = \underbrace{\int x^2 d \left(\dfrac{e^{-3x}}{-3}\right)}_{\displaystyle \int v du} = \overbrace{\dfrac{e^{-3x}}{-3} x^2}^{\displaystyle uv} - \underbrace{\int \dfrac{e^{-3x}}{-3} (2x) dx}_{\displaystyle \int u dv} = \dfrac{e^{-3x}}{-3} x^2 + \dfrac23 \int xe^{-3x} dx$$ Now let us do integration by parts again to evaluate $\displaystyle \int xe^{-3x} dx$. $$\int xe^{-3x} dx = \underbrace{\int x d \left(\dfrac{e^{-3x}}{-3}\right)}_{\displaystyle \int v du} = \overbrace{x \dfrac{e^{-3x}}{-3}}^{\displaystyle uv} - \underbrace{\int \dfrac{e^{-3x}}{-3} dx}_{\displaystyle \int u dv} = -\dfrac{x e^{-3x}}3 - \dfrac{e^{-3x}}{27}$$ Putting this back, we get that $$\int x^2 e^{-3x} dx = -\dfrac13 e^{-3x}x^2 - \dfrac29 e^{-3x}x - \dfrac2{27} e^{-3x} + c = - \dfrac{e^{-3x}}3 \left(x^2 + \dfrac{2x}3 + \dfrac2{27}\right) + c$$

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Generally, if you're gonna put two $$ you don't need to use \dfrac. –  Asaf Karagila May 12 '13 at 2:24
    
@AsafKaragila Ok. But why not? Is there any difference when we view it? –  user17762 May 12 '13 at 2:26
    
The d in the \dfrac is short for \displaystyle, so whenever you write \dfrac{}{} it is equivalent to writing {\displaystyle\frac{}{}}. Since in $$ mode you already find yourself in a \displaystyle environment, there's no need to use \dfrac again. Similarly if you have explicitly stated \displaystyle in the code prior to the \frac's. –  Asaf Karagila May 12 '13 at 2:28

Hold onto your hat: For the integral $$\int x^2e^{-3x}dx$$ we will let $u= x^2$ and $dv=e^{-3x}dx$. Thus, $du = 2x dx$ and $v = \dfrac{e^{-3x}}{-3}$.

Thus we have $$\int x^2e^{-3x}dx = x^2 \frac{e^{-3x}}{-3} - \int 2x \dfrac{e^{-3x}}{-3} dx.$$

For the latter integral above, we simplify and factor out the constants so that we have $$\dfrac{2}{-3} \int x e^{-3x}dx.$$

Now we let $u= x$, $dv = e^{-3x}dx$. Then $du=dx, v=\dfrac{e^{-3x}}{-3}$. Hence we have $$\dfrac{2}{-3} \int x e^{-3x}dx = \dfrac{2}{3} \left( x \dfrac{e^{-3x}}{-3} - \int \dfrac{e^{-3x}}{-3} dx\right) = \dfrac{2}{-3} \left(x \dfrac{e^{-3x}}{-3} + \dfrac{e^{-3x}}{-9} \right).$$

Putting this all together,

$$\int x^2e^{-3x}dx = e^{-3x} \left(\frac{1}{-3} x^2 - \frac{2}{9}x -\frac{2}{27} \right) + C$$

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