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Explore Stack InternalAnother method can be done by substitution as: Substitite $(x-1) = u^2$ That gives $ x = u^2+1$ And $$dx = 2udu$$ The integral becomes $\int {2 u^2(u^2+1)} du $ And its integration is $ \frac {u^3}{15} [3u^2+5] +C$$ \frac {2u^3}{15} [3u^2+5] +C$ After substituting the value of $u^2=x-1$ you get $$\frac{ (x-1)^{\frac{3}{2}}}{15}[3x+2] +C $$$$\frac{ 2(x-1)^{\frac{3}{2}}}{15}[3x+2] +C $$
Another method can be done by substitution as: Substitite $(x-1) = u^2$ That gives $ x = u^2+1$ And $$dx = 2udu$$ The integral becomes $\int {2 u^2(u^2+1)} du $ And its integration is $ \frac {u^3}{15} [3u^2+5] +C$ After substituting the value of $u^2=x-1$ you get $$\frac{ (x-1)^{\frac{3}{2}}}{15}[3x+2] +C $$
Another method can be done by substitution as: Substitite $(x-1) = u^2$ That gives $ x = u^2+1$ And $$dx = 2udu$$ The integral becomes $\int {2 u^2(u^2+1)} du $ And its integration is $ \frac {2u^3}{15} [3u^2+5] +C$ After substituting the value of $u^2=x-1$ you get $$\frac{ 2(x-1)^{\frac{3}{2}}}{15}[3x+2] +C $$
Another method can be done by substitution as: Substitite $(x-1) = u^2$ That gives $ x = u^2+1$ And $$dx = 2udu$$ The integral becomes $\int {2 u^2(u^2+1)} du $ And its integration is $ \frac {u^3}{15} [3u^2+5] +C$ After substituting the value of $u^2=x-1$ you get $$\frac{ (x-1)^{\frac{3}{2}}}{15}[3x+2] +C $$
