Evaluate $ \displaystyle\int_{a}^{b}\frac{\ln{x}}{x^2+ab}\:\mathrm{d}x$, where $0 Nothing meaningful came to my mind.
Evaluate $ \int_{a}^{b}\frac{\ln{x}}{x^2+ab}dx$
2
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real-analysis
integration
definite-integrals
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0i think for such a integral exists no solution in the known elementary functions – 2017-02-07
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0@Dr.SonnhardGraubner: that is false. – 2017-02-07
1 Answers
10
It is useful to exploit a symmetry. If we replace $x$ with $\frac{ab}{x}$ we get: $$ I = \int_{a}^{b}\frac{\log x}{x^2+ab}\,dx = \int_{a}^{b}\frac{\log(ab)-\log(x)}{x^2+ab}\,dx $$ hence: $$ 2\, I = \log(ab)\int_{a}^{b}\frac{dx}{x^2+ab} = \frac{\log(ab)}{\sqrt{ab}}\left(\arctan\sqrt{\frac{b}{a}}-\arctan\sqrt{\frac{a}{b}}\right)$$ and $$\boxed{ \int_{a}^{b}\frac{\log x}{x^2+ab}\,dx = \color{red}{\frac{\log(ab)}{2\sqrt{ab}}\,\arctan\left(\frac{b-a}{2\sqrt{ab}}\right)}.}$$
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2whenever I see the question on integral, I am waiting for your answer,+1) – 2017-02-07