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I can not understand that how it is proved, so please somebody help me.

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    Start by posing a self-contained problem in the body of your Question. Relying wholly on the title to carry the burden of problem statement invites confusion on the part of Readers, and leads you to omit what context you found the problem to pose. – 2017-02-12

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$$\frac d{dx} \left[\log (x+1)\right] = \frac 1{x+1} = 1-x+x^2-x^3+\cdots$$. Now integrate termwise.

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    Plese help to give ditails – 2017-02-12
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    That's as detailed add it gets without giving a step-by-step solution. – 2017-02-12
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    Do you know how to generate a Maclaurin series? – 2017-02-12
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    @SarojGhosh $\frac d{dx}\log(x+1) = 1-x+x^2-x^3+\cdots \implies \int \frac{d}{dx}\log(x+1) dx = \int 1 - x + x^2 - x^3+\cdots \, dx$ – 2017-02-12
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    @displayname I dont know that how to generate this – 2017-02-12