What is the integral of (x^2+3x+1) ?

The integral of (x^2+3x+1) is (x^3)/3+(3x^2)/2+x+C

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integral of (x^2+3x+1)

\int (x^{2} + 3 x + 1) d x

\frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} + x + C

Solution steps

\int x^{2} + 3 x + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x + \int 3 x d x + \int 1 d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 1 d x = x

= \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} + x

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} + x + C

\frac{d y}{d x} + (x^{2} - 1) y = - x y^{6}

\int 0.03 e^{- 0.03 x} d x

\frac{\partial}{\partial y} (arctan (yz))

e^{- x} (y^{^{'}} - y) = y^{2}

x \to \infty lim (\frac{ln ( 1 + e ^{- x} )}{e ^{- x}})

What is the integral of (x^2+3x+1) ?
The integral of (x^2+3x+1) is (x^3)/3+(3x^2)/2+x+C