integration

[sujith]

Please help me with this sum.....in image

 

[mmm4444bot]

What have you tried or thought about, so far?

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Cheers :cool:

 

[HallsofIvy]

Do you know the product rule: (fg)'= f'g+ fg' ?

Do you know that the derivative of x+ a, for any a, is 1?

Do you know that the derivative of ln|x+1| is 1/(x+ 1)?

Those are what you need to do this problem.

 

[sujith]

Do you know the product rule: (fg)'= f'g+ fg' ?

Do you know that the derivative of x+ a, for any a, is 1?

Do you know that the derivative of ln|x+1| is 1/(x+ 1)?

Those are what you need to do this problem.

Yes I know the product rule integration of the first part is not tough I got stuck on not recognizing the. Way for the third part of the question.

 

[HallsofIvy]

Yes I know the product rule integration of the first part is not tough I got stuck on not recognizing the. Way for the third part of the question.

I presume you mean "product rule differentiation". Do you also know, then, that \(\displaystyle ln\left(\sqrt[7]{\frac{\pi}{|x+ 1|}}\right)= \frac{1}{7}ln(\pi)- \frac{1}{7}ln|x+ 1|\)? Now, what did you get for the first part of the problem and how does that apply to the third part?

 

[sujith]

I presume you mean "product rule differentiation". Do you also know, then, that \(\displaystyle ln\left(\sqrt[7]{\frac{\pi}{|x+ 1|}}\right)= \frac{1}{7}ln(\pi)- \frac{1}{7}ln|x+ 1|\)? Now, what did you get for the first part of the problem and how does that apply to the third part?

All this time I was thinking to substitute something for x+1 ... Now this sum is easy because the integral of lnlx+1l is obtained in part one...that's simple now...