# 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!

2 years, 12 months ago

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Comment deleted Jun 28, 2015

Sir, Ur Solution is wrong. Please check it again. There are a lot of mistakes.

- 2 years, 12 months ago

How did you get from the second last step to the last step,or more precisely how did you integrate those functions?

- 2 years, 12 months 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.

- 2 years, 11 months ago

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

- 2 years, 12 months ago

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

- 2 years, 12 months ago

By the way really nice solution!

- 2 years, 12 months ago

Thank You sir

- 2 years, 12 months ago

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

- 2 years, 12 months ago

Yes sir U are "sir" for me.

- 2 years, 12 months ago

Comment deleted Jun 29, 2015

- 2 years, 11 months ago

Sir , I don't see any. Can U please tell ?

- 2 years, 11 months ago

oh sorry, on looking closely it shows $$\frac{1}{a+1}$$ otherwise i thought u had written $$\frac{1}{a-1}$$

- 2 years, 11 months ago

Wow!Thanx for your solution!You are a genius!

- 2 years, 12 months ago

Thanks ! I am removing this solution.(as you have seen it)

- 2 years, 11 months ago

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

- 2 years, 12 months ago

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

- 2 years, 12 months 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.

- 2 years, 12 months ago

No problem bhaiya!I understand!

- 2 years, 12 months ago