What is the integral of (3-2t+t^2) ?

The integral of (3-2t+t^2) is 3t-t^2+(t^3)/3+C

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integral of (3-2t+t^2)

\int (3 - 2 t + t^{2}) d t

3 t - t^{2} + \frac{t ^{3}}{3} + C

Solution steps

\int 3 - 2 t + t^{2} d t

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 d t - \int 2 t d t + \int t^{2} d t

\int 3 d t = 3 t

\int 2 t d t = t^{2}

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

= 3 t - t^{2} + \frac{t ^{3}}{3}

Add a constant to the solution

= 3 t - t^{2} + \frac{t ^{3}}{3} + C

\int \frac{sin ( x )}{5 + cos ( x )} d x

\frac{\partial}{\partial x} (x^{2} - x yz)

y^{^{''}} - 3 y^{^{'}} - 4 y = - 8 e^{x} cos (2 x)

\int (e^{y^{2}}) d y

y = 3 x

What is the integral of (3-2t+t^2) ?
The integral of (3-2t+t^2) is 3t-t^2+(t^3)/3+C