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

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

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integral of (3x+1)^2

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

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

Solution steps

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

Expand

(3 x + 1)^{2} : 9 x^{2} + 6 x + 1

= \int 9 x^{2} + 6 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 9 x^{2} d x + \int 6 x d x + \int 1 d x

\int 9 x^{2} d x = 3 x^{3}

\int 6 x d x = 3 x^{2}

\int 1 d x = x

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

Add a constant to the solution

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

e^{- 0.5 x}

\frac{d}{d t} ((A t^{2} + Bt + C) (sin (2 t) + cos (2 t)))

x \to \infty lim (\frac{x}{r})

f (x) = \frac{x ^{2} - 1}{x + 1}

\frac{d}{d x} (\frac{2}{5} x^{5} - 4 x^{3} + 9 x^{2} - 6 x - 2)