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# Anyone?

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

Note by Tanishq Varshney
2 years, 6 months ago

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Well I am getting this -

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

- 2 years, 6 months ago

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

$$\large{\displaystyle \int^{16}_{1} \frac{\ln x}{\sqrt{x}(x+4)}dx}$$

- 2 years, 6 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.

- 2 years, 6 months ago

yup plz

- 2 years, 6 months ago

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 )

- 2 years, 6 months ago

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

Thanks.

- 2 years, 6 months ago

Clever substitution!

- 2 years, 6 months ago

But I need to check once again.

- 2 years, 6 months ago