Help with the evaluation of this definite integral. I don't know where to start.
So we have :$\frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}=\frac{\ln(9-x)-\sqrt{\ln(9-x)}\sqrt{\ln(x+3)} }{\ln(9-x)-\ln(x+3)}$ What's now?
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1Use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee Nov 23 '18 at 14:30
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1https://math.stackexchange.com/questions/578957/definite-integral-int-24-frac-sqrt-log9-x-sqrt-log9-x-sqrt-log3/578960#578960 – lab bhattacharjee Nov 23 '18 at 14:33
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1Also https://math.stackexchange.com/questions/957510 – Nosrati Nov 23 '18 at 14:36
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1https://math.stackexchange.com/questions/1073120/integral-int-12011-frac-sqrtx-sqrt2012-x-sqrtxdx – lab bhattacharjee Nov 23 '18 at 14:36
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$\displaystyle \int_{2}^{4}\frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}=I=\int_{2}^{4}\frac{\sqrt{\ln(x+3)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}$
$\displaystyle\implies 2I=\int_{2}^{4}\frac{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} \:dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}=\int_{2}^{4} dx=2\implies I=1.$
$\left(\text{Using }\:\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\right)$.
Yadati Kiran
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