int frac{sqrt{1-x^2} sin ^{-1} x+x}{sqrt{1-x^ | vedclass

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Question

(C) આપેલ છે,$\int \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x+x}{\sqrt{1-x^2}} d x$ $= \int \left( \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x}{\sqrt{1-x^2}} + \f

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$x \sin ^{-1} x+c$

Vedclass · Answer

(C) આપેલ છે,$\int \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x+x}{\sqrt{1-x^2}} d x$ $= \int \left( \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}} \right) d x$ $= \int \left( \sin ^{-1} x + \frac{x}{\sqrt{1-x^2}} \right) d x$ $= \int \sin ^{-1} x d x + \int \frac{x}{\sqrt{1-x^2}} d x$ $\int \sin ^{-1} x d x$ માટે ખંડશઃ સંકલનનો ઉપયોગ કરતા,જ્યાં $f(x) = \sin ^{-1} x$ અને $g(x) = 1$: $= \sin ^{-1} x \cdot x - \int \left( \frac{d}{d x} \sin ^{-1} x \cdot \int 1 d x \right) d x + \int \frac{x}{\sqrt{1-x^2}} d x$ $= x \sin ^{-1} x - \int \frac{1}{\sqrt{1-x^2}} \cdot x d x + \int \frac{x}{\sqrt{1-x^2}} d x + c$ $= x \sin ^{-1} x + c$ તેથી,વિકલ્પ $(C)$ સાચો છે.

$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$

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