Prove that,$\int_{0}^{\infty}\dfrac{1}{x^3}\sin^3\left(x-\dfrac{1}{x} \right)^5\, dx=0$
and$\int_{-1}^{0}\dfrac{x^2+2x}{\ln{(x+1)}} dx=\ln{3}$.I have no idea how to solve these problems!

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@Adarsh Kumar @Rajdeep Dhingra The first Integral is really hard. Also, it is not equal to $0.$ The Integral is equivalent to $-\frac15 \int_0^\infty \sin^3(u) u^{-3/5} du \approx {-0.1535}$ Anyway, to be able to compute the original Integral, you need to know some conditions for which an Integral becomes Riemann Improper Integrable. I can try to proceed if you want, but this is really miles and miles above our level.

Hello Adarsh. I'm really sorry but I won't be able to help now; I'm truly exhausted after coaching class since 10 in the morning - and the same schedule is in place for tomorrow. I'll answer as soon as I can for sure.

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TopNewest@Adarsh Kumar @Rajdeep Dhingra The first Integral is really hard. Also, it is not equal to $0.$ The Integral is equivalent to $-\frac15 \int_0^\infty \sin^3(u) u^{-3/5} du \approx {-0.1535}$ Anyway, to be able to compute the original Integral, you need to know some conditions for which an Integral becomes Riemann Improper Integrable. I can try to proceed if you want, but this is really miles and miles above our level.

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@Adarsh Kumar Seems you've got some excellent answers. I would have done the second the same way as Rajdeep.

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Sir the 1st one ? @Samarpit Swain solution is wrong.

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By the way really nice solution!

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Come on Rajdeep! I'm not Sir!$$I'll have a go at it in half an hour.

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Hint about second integral, take x+1=t, and do it with the help of differentiation under integral.

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Actually i tried that and I got,$\int_{0}^{1}\dfrac{t^2-1}{\ln{t}}dt$,how to proceed now?

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Hello Adarsh. I'm really sorry but I won't be able to help now; I'm truly exhausted after coaching class since 10 in the morning - and the same schedule is in place for tomorrow. I'll answer as soon as I can for sure.

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No problem bhaiya!I understand!

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@Nihar Mahajan @Calvin Lin @Ronak Agarwal @Rajdeep Dhingra @Ishan Dasgupta Samarendra @Nishant Rai

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