જો int x(1+x) log(1+x^2) dx = F(x) log(1+x^2) | vedclass

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(C) ધારો કે $I = \int (x+x^2) \log(1+x^2) dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા,$\log(1+x^2)$ ને પ્રથમ વિધેય લેતા:$I = \log(1+x^2) \int (x+x^2) dx - \int \lef

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$\frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2}$

Vedclass · Answer

(C) ધારો કે $I = \int (x+x^2) \log(1+x^2) dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા,$\log(1+x^2)$ ને પ્રથમ વિધેય લેતા:$I = \log(1+x^2) \int (x+x^2) dx - \int \left( \frac{d}{dx} \log(1+x^2) \int (x+x^2) dx ight) dx$$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} ight) - \int \frac{2x}{1+x^2} \left( \frac{x^2}{2} + \frac{x^3}{3} ight) dx$$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} ight) - \int \left( \frac{x^3}{1+x^2} + \frac{2x^4}{3(1+x^2)} ight) dx$સંકલનનું મૂલ્ય શોધતા:$\int \frac{x^3}{1+x^2} dx = \int \frac{x(x^2+1-1)}{1+x^2} dx = \int x dx - \int \frac{x}{1+x^2} dx = \frac{x^2}{2} - \frac{1}{2} \log(1+x^2)$$\int \frac{2x^4}{3(1+x^2)} dx = \frac{2}{3} \int \frac{x^4-1+1}{1+x^2} dx = \frac{2}{3} \int (x^2-1) dx + \frac{2}{3} \int \frac{1}{1+x^2} dx = \frac{2x^3}{9} - \frac{2x}{3} + \frac{2}{3} 	an^{-1} x$કિંમતો મૂકતા:$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} ight) - \left( \frac{x^2}{2} - \frac{1}{2} \log(1+x^2) + \frac{2x^3}{9} - \frac{2x}{3} + \frac{2}{3} 	an^{-1} x ight) + c$$I = \log(1+x^2) \left( \frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2} ight) - \frac{2}{3} 	an^{-1} x - \frac{2x^3}{9} - \frac{x^2}{2} + \frac{2x}{3} + c$આપેલ સમીકરણ સાથે સરખાવતા,$F(x) = \frac{x^2}{2} + \frac{x^3}{3} + \frac{1}{2}$ મળે છે.

જો $\int x(1+x) \log(1+x^2) dx = F(x) \log(1+x^2) - \frac{2}{3} \tan^{-1} x - \frac{2x^3}{9} - \frac{x^2}{2} + \frac{2x}{3} + c$ હોય,તો $F(x) =$

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