# Integral

• Oct 2nd 2009, 06:40 PM
pankaj
Integral
Evaluate
$\displaystyle \int_{0}^1\frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{(1-(1+x^2)\cot^{-1}x\}}dx$
• Oct 3rd 2009, 12:43 AM
NonCommAlg
Quote:

Originally Posted by pankaj
Evaluate
$\displaystyle \int_{0}^1 \frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{1-(1+x^2)\cot^{-1}x\}}dx$

man, this is one ugly integral! haha ...

ok, we have this identity: $\displaystyle \frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{1-(1+x^2)\cot^{-1}x\}}=1 + \frac{2x}{1+x^2} - \frac{1-2x \cot^{-1} x}{1-(1+x^2)\cot^{-1}x}.$ we also have $\displaystyle \frac{d}{dx}(1 - (1+x^2) \cot^{-1}x) = 1 - 2x \cot^{-1} x.$

thus: $\displaystyle \int \frac{2x-(1+x^2)^2\cot^{-1}x}{(1+x^2)\{1-(1+x^2)\cot^{-1}x\}} \ dx=x + \ln \left|\frac{1+x^2}{1 - (1+x^2)\cot^{-1} x} \right| + C.$ the value of your definite integral is $\displaystyle 1+\ln 2.$
• Oct 4th 2009, 01:07 AM
simplependulum
Quote:

Originally Posted by NonCommAlg
man, this is one ugly integral! haha ...

I am hungry for your beautiful integrals ....
• Oct 4th 2009, 05:42 AM
pankaj
Here is my solution:

$\displaystyle \int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx$

=$\displaystyle \int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\frac{1}{1+x^2}+\frac{1}{1+x^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx$

=$\displaystyle -\int_{0}^1\frac{\frac{-2x}{(1+x^2)^2}+\frac{1}{1+x^2}}{\frac{1}{1+x^2}-\cot^{-1}x}dx+\int_{0}^11dx$

=$\displaystyle \ln 2 +1$
• Oct 4th 2009, 08:32 PM
NonCommAlg
Quote:

Originally Posted by pankaj
Here is my solution:

$\displaystyle \int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx$

=$\displaystyle \int_{0}^1\frac{\frac{2x}{(1+x^2)^2}-\frac{1}{1+x^2}+\frac{1}{1+x^2}-\cot^{-1}x}{\frac{1}{1+x^2}-\cot^{-1}x}dx$

=$\displaystyle -\int_{0}^1\frac{\frac{-2x}{(1+x^2)^2}+\frac{1}{1+x^2}}{\frac{1}{1+x^2}-\cot^{-1}x}dx+\int_{0}^11dx$

=$\displaystyle \ln 2 +1$

this is better than my solution! (Clapping)