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Originally Posted by khhratmath123 solve this question....... Show us your work so far and then we can guide you through where you're stuck. As a pointer, the way I would approach this is to factorise the denominator, and then use partial fractions to get two fractions which are easy to integrate.
Originally Posted by khhratmath123 solve this question....... Also, if you are having problems factorising the fraction, remember you can take a factor of -1 out, so your problem becomes: $\displaystyle (-1)\int^{-2}_{-3}\frac{1}{x^2 + 6x+5}dx$
i have solved this question but i have doubt....plz check..is it right?
this was easy: ∫((dx)/(x²+6x+5)) =∫((dx)/((x+1)(x+5))) now use partial fractions ,get the constants and put the limits. your answer is definitely wrong as it should come in terms of logs.
Originally Posted by khhratmath123 i have solved this question but i have doubt....plz check..is it right? Correct.
Originally Posted by khhratmath123 i have solved this question but i have doubt....plz check..is it right? No. I see what you have done, but your answer should be: $\displaystyle [\frac{1}{2}tanh^{-1}(\frac{x+3}{2})]^{-2}_{-3}$ Or the way that me and Pulock suggested, this way you would get the actual logs instead of the inverse hyperbolic.
Originally Posted by craig No. $\displaystyle \int\frac{dx}{(2)^2-(x+3)^2}$ Let $\displaystyle u=x+3\Rightarrow{du}=dx$ $\displaystyle \Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}$
Originally Posted by VonNemo19 $\displaystyle \int\frac{dx}{(2)^2-(x+3)^2}$ Let $\displaystyle u=x+3\Rightarrow{du}=dx$ $\displaystyle \Rightarrow\int\frac{dx}{(2)^2-(x+3)^2}=\int\frac{du}{(2)^2-(u)^2}$ I see how you derived that. However, if it was: $\displaystyle \int\frac{du}{\sqrt{(2)^2-(u)^2}}$ that would make his answer right. But as far as I remember, $\displaystyle \int\frac{du}{1-u^2} = tanh^{-1} u$ ?
Originally Posted by craig I see how you derived that. However, if it was: $\displaystyle \int\frac{du}{\sqrt{(2)^2-(u)^2}}$ that would make his answer right. But as far as I remember, $\displaystyle \int\frac{du}{1-u^2} = tanh^{-1} u$ ? My mind saw a radical where there was none.
Originally Posted by VonNemo19 My mind saw a radical where there was none. Easy to do
thanks for all i have solved by partial fraction method
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