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Question

Find the indefinite integral. (2+x+2x2+ex)dx\int\left(2+x+2 x^{2}+e^{x}\right) d x

Solution

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Applying the extended version of Rule\textbf{Rule} 4\pmb{4} we get:

(2+x+2x2+ex)dx=2dx+xdx+2x2dx+exdx\int (2+x+2x^2+e^x)\, dx=\int 2\, dx+\int x\,dx+\int 2x^2\, dx+\int e^x\,dx

Applying Rule 33 on the first and the third integral we get

(2+x+2x2+ex)dx=2dx+xdx+2x2dx+exdx=2dx+xdx+2x2dx+exdx\begin{align*}\int (2+x+2x^2+e^x)\, dx&=\int 2\, dx+\int x\,dx+\int 2x^2\, dx+\int e^x\,dx\\ &=2\int \, dx+\int x\, dx+2\int x^2\, dx+\int e^x\, dx \end{align*}

Finally, applying Rule 55 and Rule 22 we have that

(2+x+2x2+ex)dx=2x+x22+2x33+ex+C\boxed{\int(2+x+2x^2+e^x)\, dx=2x+\frac{x^2}{2}+2\cdot \frac{x^3}{3}+e^x+C}

Where Rule\textbf{Rule} 4\pmb{4}, Rule\textbf{Rule} 5\pmb{5} , Rule\textbf{Rule} 2\pmb{2} and Rule\textbf{Rule} 3\pmb{3} are given by

[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}

exdx=ex+C\pmb{\int e^x\, dx=e^x+C}

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

cf(x)dx=cf(x)dx\pmb{\int cf(x)\, dx=c\int f(x)\, dx}

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