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Saturday, 13 August 2011

Math GRE - #12

A simple one today:
What is the value of the integral: \[\int_0^1{\frac{x}{1+x^2}}dx?\]
  • $1$
  • $\frac{\pi}{4}$
  • $\arctan\frac{\sqrt{2}}{2}$
  • $\log 2$
  • $\log\sqrt{2}$

Solution :

We can use the substitution $u=1+x^2\implies du = 2x\,dx$. With this substitution, the original integral becomes: \[\int_0^1{\frac{x}{1+x^2}}dx=\frac{1}{2}\int_1^2{\frac{1}{u}}du\]
Upon computing this integral, we obtain: \[\frac{1}{2}\int_1^2{\frac{1}{u}}du=\frac{1}{2}\left. \log{u}\right|_1^2=\frac{1}{2}\log{2}=\log{\sqrt{2}}.\]


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