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\(\large{\displaystyle \int^{c}_{b} \frac{\ln x}{\sqrt{x}(x+a)}dx}\)

Note by Tanishq Varshney
2 years ago

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Well I am getting this -

\(\displaystyle \dfrac{1}{a} \left[\frac{\sqrt{z} log(z)}{z + a} + \frac{\sqrt{z} log(z)}{2a} \Phi\left(-\frac{z}{a}, 1, \frac{1}{2}\right) - \frac{\sqrt{z}}{a} \Phi\left(-\frac{z}{a}, 1, \frac{1}{2}\right)\right]_b^c\) Kartik Sharma · 2 years ago

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@Kartik Sharma ok frankly i didnt understand what u have typed here. anyway i was trying to find

\(\large{\displaystyle \int^{16}_{1} \frac{\ln x}{\sqrt{x}(x+4)}dx}\) Tanishq Varshney · 2 years ago

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@Tanishq Varshney Substitute \( x = y^2 \). Then put \(y=\dfrac{2}{t} \) to again get the same integral and evaluate it. I'll post the complete solution if you want. Sudeep Salgia · 2 years ago

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@Sudeep Salgia yup plz Tanishq Varshney · 2 years ago

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@Tanishq Varshney Let \( \displaystyle I = \int_1^{16} \frac{ \ln x }{\sqrt{x}(x+4) } dx \). Substitute \( x = y^2 \) to get \( \displaystyle I = \int_1^{4} \frac{ 4 \ln y }{(y^2+4) } dy \). Now put \( y = \dfrac{4}{t} \) to get \( \displaystyle \frac{I}{4} = \int_1^4 \frac{ \ln 4 - \ln t}{(4 + t^2)} dt \). Rearranging, we get, \( \displaystyle I = 2 \ln 4 \int_1^4 \frac{1}{t^2 + 4} dt \).
This is a simple integration. So the answer becomes, \( I = \ln 4 ( \arctan (2) - \arctan(1/2)) \)

Hope this helps. Please check and let me know if there are any calculation mistakes. (I wasn't writing any thing down :P ) Sudeep Salgia · 2 years ago

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@Sudeep Salgia Can you please explain the method where you divide I/4 . And when you re-arrange ln (t) disappears ?

Thanks. Syed Baqir · 2 years ago

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@Sudeep Salgia Clever substitution! Aditya Kumar · 2 years ago

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@Kartik Sharma But I need to check once again. Kartik Sharma · 2 years ago

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