What is the integral of (3x^2+4x+1) ?

The integral of (3x^2+4x+1) is x^3+2x^2+x+C

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integral of (3x^2+4x+1)

\int (3 x^{2} + 4 x + 1) d x

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

Solution steps

\int 3 x^{2} + 4 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 3 x^{2} d x + \int 4 x d x + \int 1 d x

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

\int 4 x d x = 2 x^{2}

\int 1 d x = x

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

Add a constant to the solution

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

n = 0 \sum \infty (- x^{2} - 8)^{n}

\int \frac{1}{2} e^{x^{2}} d x

(x^{2} - 4) \cdot \frac{d y}{d x} + 4 y = (x + 2)^{2}

\int 12 x y d x

x^{2} y^{^{''}} - x y^{^{'}} + 2 y = 0

What is the integral of (3x^2+4x+1) ?
The integral of (3x^2+4x+1) is x^3+2x^2+x+C