Can somebody help me integrate r(t)=t/1+t^2 I use the substitution method t(1+t^2)^-1 let 1+t^2=u 2tdt=du dt=du/2t hence putting in ques r(t)=∫t/u^-1 du/2t =1/2∫u^-1 du =1/2u^0/0 which is not possible .Any help?
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What is the derivative of $\displaystyle 1+t^{2}$?... the answer is also the solution of integral ... Merry Christmas from Italy $\displaystyle \chi$ $\displaystyle \sigma$
the derivative is 2tdt .But I did this at the top and I end up with some other number.
Well, what's $\displaystyle \int \frac{1}{x}dx$ ?
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