What is the integral of (3t^2-4t+5) ?

The integral of (3t^2-4t+5) is t^3-2t^2+5t+C

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integral of (3t^2-4t+5)

\int (3 t^{2} - 4 t + 5) d t

t^{3} - 2 t^{2} + 5 t + C

Solution steps

\int 3 t^{2} - 4 t + 5 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 t^{2} d t - \int 4 t d t + \int 5 d t

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

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

\int 5 d t = 5 t

= t^{3} - 2 t^{2} + 5 t

Add a constant to the solution

= t^{3} - 2 t^{2} + 5 t + C

\int sin^{4} (x) cos^{2} (d) x d x

\frac{\partial}{\partial y} ((1 + x y)^{\frac{3}{2}})

\int \frac{cos ( x )}{cos ^{2} ( x )} d x

arctan (x^{3})

y = \frac{x ^{3} + 3 x ^{2} - 5 x + 2}{3}

What is the integral of (3t^2-4t+5) ?
The integral of (3t^2-4t+5) is t^3-2t^2+5t+C