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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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