# Integration problems, help!

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! 6 years ago

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- 6 years ago

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.

- 6 years ago

No problem bhaiya!I understand!

- 6 years ago

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

- 6 years ago

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

- 6 years ago

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

- 6 years ago

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

- 6 years ago

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

- 6 years ago

Yes sir U are "sir" for me.

- 6 years ago

By the way really nice solution!

- 6 years ago

Thank You sir

- 6 years ago

@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.

- 6 years ago