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Question

Find the indefinite integral. (1+u+u2)du\int\left(1+u+u^{2}\right) d u

Solution

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Applying Rule\textbf{Rule},,4\pmb{4} we get

(1+u+u2)du=du+udu+u2du=u+u1+11+1+u2+12+1+C\begin{align*} \displaystyle\int{(1+u+u^2)\, du}&= \displaystyle\int{\, du}+\displaystyle\int{u\, du}+\displaystyle\int{u^2\, du}\tag{\textcolor{#c34632}{Rule $2$}}\\ &=u+\frac{u^{1+1}}{1+1}+\frac{u^{2+1}}{2+1}+C\\ \end{align*}

Finally, we get

(1+u+u2)du=u+u22+u33+C\boxed{\boxed{\displaystyle\int{(1+u+u^2)\, du=u+\frac{u^2}{2}+\frac{u^3}{3}+C}}}

Where $\textbf{Rule 22}$ and $\textbf{Rule 44}$ are given by

xndx=1n+1xn+1+C\pmb{\int{x^n}\, dx=\frac{1}{n+1}\cdot x^{n+1}+C}

[f(x)±g(x)]dx=f(x)dx±g(x)dx\pmb{\int[{f(x)\pm g(x)]\, dx}=\int{f(x)}\, dx\pm\int{g(x)\, dx}}

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