NTA Abhyas 2020 · Mathematics · Indefinite Integration
4
The given integral is I=∫x3−1(x+1)(x4+1)dxI = ∫ \dfrac{{x}^{3} - 1}{\left(x + 1\right)\left({x}^{4} + 1\right)}dxI=∫(x+1)(x4+1)x3−1dx=∫x3+x4−x4−1(x+1)(x4+1)dx= ∫ \dfrac{{x}^{3} + {x}^{4} - {x}^{4} - 1}{\left(x + 1\right)\left({x}^{4} + 1\right)}dx=∫(x+1)(x4+1)x3+x4−x4−1dx=∫x3(x+1)−(x4+1)(x+1)(x4+1)dx= ∫ \dfrac{{x}^{3}\left(x + 1\right) - \left({x}^{4} + 1\right)}{\left(x + 1\right)\left({x}^{4} + 1\right)}dx=∫(x+1)(x4+1)x3(x+1)−(x4+1)dx=∫x3x4+1dx−∫dxx+1= ∫ \dfrac{{x}^{3}}{{x}^{4} + 1}dx - ∫ \dfrac{dx}{x + 1}=∫x4+1x3dx−∫x+1dx=14∫4x3dxx4+1−∫dxx+1= \dfrac{1}{4}∫ \dfrac{4{x}^{3}dx}{{x}^{4} + 1} - ∫ \dfrac{dx}{x + 1}=41∫x4+14x3dx−∫x+1dx=14ln(x4+1)−ln(1+x)+c= \dfrac{1}{4}\ln \left({x}^{4} + 1\right) - \ln \left(1 + x\right) + c=41ln(x4+1)−ln(1+x)+c
Hence, f(x)=x4+1,g(x)=x+1f\left(x\right) = {x}^{4} + 1, g\left(x\right) = x + 1f(x)=x4+1,g(x)=x+1∴f(1)⋅g(1)=2×2=4∴ f\left(1\right) \cdot g\left(1\right) = 2 \times 2 = 4∴f(1)⋅g(1)=2×2=4
