@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.
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Sudeep Salgia
·
1 year, 8 months ago

@Tanishq Varshney
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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
·
1 year, 8 months 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 ?

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TopNewestWell 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 · 1 year, 8 months ago

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\(\large{\displaystyle \int^{16}_{1} \frac{\ln x}{\sqrt{x}(x+4)}dx}\) – Tanishq Varshney · 1 year, 8 months ago

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– Sudeep Salgia · 1 year, 8 months ago

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.Log in to reply

– Tanishq Varshney · 1 year, 8 months ago

yup plzLog in to reply

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 · 1 year, 8 months ago

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Thanks. – Syed Baqir · 1 year, 8 months ago

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– Aditya Kumar · 1 year, 8 months ago

Clever substitution!Log in to reply

– Kartik Sharma · 1 year, 8 months ago

But I need to check once again.Log in to reply