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I'm practicing to solve a whole, and I am not able to solve this one, could you help me? $$\int \frac{dt}{(t+2)^2(t+1)}$$I tried $$\frac{1}{(t+2)^2(t+1)}=\frac{A}{(t+2)^2}+\frac{B}{(t+2)}+\frac{C}{(t+1)}\\1=A(t+1)+B(t+2)(t+1)+C(t+2)^2\\t=-2\Longrightarrow 1=-A\Longrightarrow \fbox{$A=-1$}\\t=-1\Longrightarrow1=C(-1)^2\Longrightarrow\fbox{$C=1$}$$Making $t = 0$ and substituting $A$ and $C$ we have $$1=A+2B+4C=-1+2B+4=2B+3\\\fbox{$B=-1$}$$THEN$$\int \frac{dt}{(t+2)^2(t+1)}=\int -\frac{1}{(t+2)^2}-\frac{1}{(t+2)}+\frac{1}{(t+1)}\;dt\\=-\int \frac{1}{(t+2)^2}-\int\frac{1}{(t+2)}+\int\frac{1}{(t+1)}\;dt=-\int u^{-2\;}du-\ln|t+2|+\ln|t+1|\\=-\frac{u^{-1}}{-1}-\ln|t+2|+\ln|t+1|=\\\fbox{$\frac{1}{t+2}-\ln|t+2|+\ln|t+1|+c$}$$Only I could not do the derivative to "take the test", can you help me? Or contains an error in my resolution?

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seems okay according to wolf –  Santosh Linkha Sep 26 '13 at 16:30
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4 Answers 4

$$\left(\frac1{t+2}+\log\frac{t+1}{t+2}\right)'=-\frac1{(t+2)^2}+\frac{t+2}{t+1}\frac1{(t+2)^2}=$$

$$=-\frac1{(t+2)^2}+\frac1{(t+1)(t+2)}=\frac{-t-1+t+2}{(t+1)(t+2)^2}=\frac1{(t+1)(t+2)^2}\;\color\red\checkmark$$

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(+1) for the checkmark sign at the end :) –  Caran-d'Ache Sep 27 '13 at 8:21
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Differentiating $$\frac{1}{t+2} - \ln |t + 2| + \ln|t+1| + c$$ gives you $$-\frac{1}{(t+2)^2} - \frac{1}{t + 2} + \frac{1}{t+1}$$ which is certainly what you got when you split the fraction initially. To check that your A, B and C were correct, see that this is just $$\frac{-(t + 1) - (t + 2)(t+1) + (t+2)^2}{(t+2)^2(t+1)}.$$ Simplifying, we get $$-(t+1)-(t+2)(t+1)+(t+2)^2 = -(t+1) - (t^2 + 3t + 2) + (t^2 + 4t +4) = 1,$$ as required.

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$$D_t\;\;\frac{1}{t+1}-\ln|t+2|+\ln|t+1|+c=\\=\frac{(t+2)\cdot0-1\cdot1}{(t+1)^2}-\frac{1}{t+2}+\frac{1}{t+1}+0=\\=\frac{-(t+1)-(t+1)(t+2)+(t+2)^2}{(t+2)^2(t+1)}=\\=\frac{-t-1-t^2-3t-2+t^2+4t+4}{(t+2)^2(t+1)}=\\=\frac{1}{(t+2)^2(t+1)}$$ Correct .. -

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\begin{align} \int{{\rm d}t \over \left(t + \mu\right)\left(t + 1\right)} &= {1 \over 1 - \mu} \int\left({1 \over t + \mu} - {1 \over t + 1}\right)\,{\rm d}t = {\ln\left(t + \mu\right) \over 1 - \mu} - {\ln\left(t + 1\right) \over 1 - \mu}\,{\rm d}t \\[1cm]& \mbox{Derive both members respect of}\ \mu \\[3mm] -\int{{\rm d}t \over \left(t + \mu\right)^{2}\left(t + 1\right)} &= {1 \over \left(t + \mu\right)\left(1 - \mu\right)} + {\ln\left(t + \mu\right) \over \left(1 - \mu\right)^{2}} - {\ln\left(t + 1\right) \over \left(1 - \mu\right)^{2}} \end{align} Set $\mu = 2$ and changes sign in both members: $$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \int{{\rm d}t \over \left(t + 2\right)^{2}\left(t + 1\right)} = {1 \over t + 2} - \ln\left(t + 2\right) + \ln\left(t + 1\right)\quad} \\ \\ \hline \end{array} $$

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