If ⨍(x)=1-x+x2-x3+x4+….∞ then ∫0xf(x)dx=

f(x)=1-x+x2-x3+.... =11-(-x)=11+x∫0x f(x) dx=∫0x 11+x dx =log (1+x)

If ⨍(x)=1-x+x2-x3+x4+….∞ then ∫0xf(x)dx=

f(x)=1-x+x2-x3+.... =11-(-x)=11+x∫0x f(x) dx=∫0x 11+x dx =log (1+x)

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If ${⨍(x)=1-x+x}^{2} {-x}^{3} {+x}^{4} +….∞$ then $\int_{0}^{x} f (x) dx =$

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$logx$

$\log (1 - x)$

$\log (1 + x)$

$\log (1 - x^{2})$

answer is C.

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$\begin{array}{rcl} f (x) & = & 1 - x + x^{2} - x^{3} + . . . . \\ = & \frac{1}{1 - (- x)} = \frac{1}{1 + x} \\ \int_{0}^{x} f (x) dx & = & \int_{0}^{x} \frac{1}{1 + x} dx \\ = & \log (1 + x) \end{array}$

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If ⨍(x)=1-x+x2-x3+x4+….∞ then ∫0xf(x)dx=

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If ${⨍(x)=1-x+x}^{2} {-x}^{3} {+x}^{4} +….∞$ then $\int_{0}^{x} f (x) dx =$